Contoh Soal 3 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 11.&(\textbf{SPMB 04})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{27}{3^{2x-1}}=81^{-0,125} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1\displaystyle \frac{3}{4}\\ \textrm{b}.&-\displaystyle \frac{3}{4}\\ \textrm{c}.&\displaystyle \frac{3}{4}\\ \textrm{d}.&1\displaystyle \frac{1}{4}\\ \color{red}\textrm{e}.&2\displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\\ 3^{3-(2x-1)}&=3^{4(\frac{1}{8})}\\ 3-2x+1&=-\displaystyle \frac{1}{2}\\ -2x+4&=-\displaystyle \frac{1}{2}\\ -x+2&=-\displaystyle \frac{1}{4}\\ -x&=-2-\displaystyle \frac{1}{4}\\ -x&=-2\displaystyle \frac{1}{4}\\ x&=2\displaystyle \frac{1}{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&(\textbf{UMPTN 98})\textrm{Bentuk}\: \: \left ( \displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^{2}} \right )^{-\frac{3}{4}}\\ &\textrm{dapat diserdernakan menjadi}\\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{xy^{2}}\\ \textrm{b}.&x\sqrt{y}\\ \textrm{c}.&\sqrt{x^{2}y}\\ \color{red}\textrm{d}.&xy\sqrt{y}\\ \textrm{e}.&xy\sqrt{x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^{2}} \right )^{-\frac{3}{4}}&=\left ( x^{\frac{2}{3}-2}.y^{-\frac{3}{4}-\frac{2}{3}} \right )^{-\frac{3}{4}}\\ &=x^{-\frac{3}{4}(\frac{2}{3}-2)}.y^{-\frac{3}{4}(-\frac{3}{4}-\frac{2}{3})}\\ &=x^{-\frac{1}{2}+\frac{3}{2}}.y^{1+\frac{1}{2}}\\ &=x^{1}.y^{1\frac{1}{2}}\\ &=xy\sqrt{y} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&(\textbf{UMPTN 00})\\ &\textrm{Bentuk}\: \: \left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ &\textrm{Jika}\: \: x_{0}\: \: \textrm{memenuhi persamaan, maka nilai}\\ &1-\displaystyle \frac{3}{4}x_{0}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\frac{3}{16}\\ \textrm{b}.&1\frac{1}{4}\\ \textrm{c}.&1\frac{3}{4}\\ \color{red}\textrm{d}.&2\frac{1}{3}\\ \textrm{e}.&2\frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left (\sqrt[3]{\displaystyle \frac{1}{243}} \right )^{3x}&=\left ( \displaystyle \frac{3}{3^{x-2}} \right )^{2}\sqrt[3]{\displaystyle \frac{1}{9}}\\ 3^{-5x}&=3^{2(1-(x-2))}.3^{-\frac{2}{3}}\\ -5x&=2(1-(x-2))+\left ( -\displaystyle \frac{2}{3} \right ),\: \: \textrm{dikali}\: \: 3\\ -15x&=6(3-x)+(-2)\\ -15x&=18-6x-2\\ 6x-15x&=16\\ -9x&=16\\ x&=\displaystyle \frac{16}{-9}\\ x_{0}&=-\displaystyle \frac{16}{9},\: \: \textrm{selanjutnya}\\ 1-\displaystyle \frac{3}{4}x_{0}&=1-\displaystyle \frac{3}{4}\times \left (-\frac{16}{9} \right )\\ &=1+\frac{4}{3}\\ &=1+1\displaystyle \frac{1}{3}\\ &=2\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Diketahui}\: \: x^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\\ &\textrm{Nilai}\: \: x+x^{-1}=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&7\\ \textrm{b}.&8\\ \textrm{c}.&8\\ \textrm{d}.&10\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}x^{\frac{1}{2}}+x^{-\frac{1}{2}}&=3\\ \textrm{dikadrat}&\textrm{kan}\\ \left ( x^{\frac{1}{2}}+x^{-\frac{1}{2}} \right )^{2}&=3^{2}\\ x+2+x^{-1}&=9\\ x+x^{-1}&=9-2\\ &=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\: \: 2^{2x}+2^{-2x}=2\\ &\textrm{Nilai}\: \: 2^{x}+2^{-x}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&2\\ \textrm{c}.&\sqrt{2}\\ \textrm{d}.&3\\ \textrm{e}.&\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}2^{2x}+2^{-2x}&=2\\ \textrm{jika soal}&\: \textrm{dikuadratkan}\\ \left ( 2^{x}+2^{-x} \right )^{2}&=2^{2x}+2+2^{-2x}\\ &=2^{2x}+2^{-2x}+2\\ \left ( 2^{x}+2^{-x} \right )^{2}&=2+2=4\\ 2^{x}+2^{-x}&=\sqrt{4}\\ &=2 \end{aligned} \end{array}$

Contoh Soal 2 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}=\displaystyle \frac{1}{25} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-5\\ \textrm{b}.&-4\\ \textrm{c}.&-3\\ \textrm{d}.&1\\ \textrm{e}.&3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}&=\displaystyle \frac{1}{25}\\ 25.\displaystyle 5^{2}\left ( \left ( \displaystyle \frac{1}{25} \right )^{2x+6} \right )^{\frac{1}{6}}&=1\\ 5^{2}.5^{2}.5^{-\frac{4x+12}{6}}&=5^{0}\\ 5^{2+2-\frac{2}{3}x-2}&=5^{0}\\ \displaystyle 2+2-\frac{2}{3}x-2&=0\\ 2-\displaystyle \frac{2}{3}x&=0\\ -\displaystyle \frac{2}{3}x&=-2\\ x&=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&(\textbf{UM-UGM 03})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\left ( \displaystyle \frac{1}{25} \right )^{x-\frac{5}{2}}=\sqrt{\displaystyle \frac{625}{5^{2-x}}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{5}\\ \color{red}\textrm{b}.&\displaystyle \frac{8}{5}\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{1}{25} \right )^{x-\frac{5}{2}}&=\sqrt{\displaystyle \frac{625}{5^{2-x}}}\\ 5^{-2x+5}&=5^{\frac{1}{2}(4-(2-x))}\\ -2x+5&=\displaystyle \frac{1}{2}(4-2+x)\\ -4x+10&=2+x\\ -5x&=2-10\\ x&=\displaystyle \frac{-8}{-5}\\ &=\frac{8}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &2^{\frac{x}{3}-1}=16 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \textrm{b}.&10\\ \color{red}\textrm{c}.&15\\ \textrm{d}.&20\\ \textrm{e}.&25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}2^{\frac{x}{3}-1}&=16\\ 2^{\frac{x}{3}-1}&=2^{4}\\ \displaystyle \frac{x}{3}-1&=4\\ \displaystyle \frac{x}{3}&=5\\ x&=15 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&(\textbf{SPMB 05})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{\sqrt[3]{(0,08)^{7-2x}}}{(0,2)^{-4x+5}}=1 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\\ \textrm{b}.&-2\\ \color{red}\textrm{c}.&-1\\ \textrm{d}.&0\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \frac{\sqrt[3]{(0,08)^{7-2x}}}{(0,2)^{-4x+5}}&=1\\ \sqrt[3]{(0,08)^{7-2x}}&=(0,2)^{-4x+5}\\ (0,2)^{\frac{3(7-2x)}{3}}&=(0,2)^{-4x+5}\\ \displaystyle 7-2x&=-4x+5\\ 4x-2x&=5-7\\ 2x&=-2\\ x&=-1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&(\textbf{SPMB 05})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\displaystyle \frac{\sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}}{27}=3^{x+1} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-16\\ \textrm{b}.&-7\\ \textrm{c}.&4\\ \textrm{d}.&5\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{\sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}}{27}&=3^{x+1}\\ \sqrt[3]{\displaystyle \frac{1}{9^{2-x}}}&=27\times 3^{x+1}\\ \sqrt[3]{3^{-2(2-x)}}&=3^{3}.3^{x+1}\\ 3^{\frac{-4+2x}{3}}&=3^{3+(x+1)}\\ \displaystyle \frac{-4+2x}{3}&=4+x\\ -4+2x&=12+3x\\ 2x-3x&=12+4\\ -x&=16\\ x&=-16 \end{aligned} \end{array}$

Contoh Soal 4 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{ll}\\ 16.&(\textbf{\textrm{UM UNDIP 2012 Math IPA}})\\ &\textrm{Jika}\: \: f(x)=x^{3}-3x^{2}-3x-1,\\ &\textrm{maka nilai dari}\: \: f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-\sqrt{2}\\ \textrm{b}.&-1\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=x^{3}-3x^{2}-3x-1\\ &=\left ( x- 1\right )^{3}-6x\\ f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )&=\left (1+\sqrt[3]{2}+\sqrt[3]{4}-1 \right )^{3}-6\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )\\ &=\left ( \sqrt[3]{2}+\sqrt[3]{4} \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left (\sqrt[3]{2} \left ( \sqrt[3]{2}+1 \right ) \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=2\left ( 1+3\sqrt[3]{2}+3\sqrt[3]{4}+2 \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left ( 6+6\sqrt[3]{2}+6\sqrt[3]{4} \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 17.&\textrm{Jika}\: \: \textbf{a}\: \: \textrm{dan}\: \: \textbf{b}\: \: \textrm{adalah bilangan bulat positif}\\ &\textrm{yang memenuhi persamaan}\: \: \textbf{a}^{\textbf{b}}=2^{20}-2^{19},\\ & \textrm{maka nilai dari}\: \: \textbf{a}+\textbf{b}=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&19\\ \color{red}\textrm{d}.&21\\ \textrm{e}.&23 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textbf{a}^{\textbf{b}}&=2^{20}-2^{19}\\ &=2^{^{^{(19+1)}}}-2^{19}\\ &=2^{19}.2^{1}-2^{19}\\ &=2^{19}(2-1)\\ &=2^{19}\\ \textrm{Se}&\textrm{hingga nilai}\: \: \textbf{a}+\textbf{b}=2+19=21\end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Perhatikan gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\quad\: \, &\textrm{Persamaan grafik fungsi seperti gambar di atas adalah}\, ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=2^{x-2}\\ \color{red}\textrm{b}.&f(x)=2^{x}-2\\ \textrm{c}.&f(x)=2^{x}-1\\ \textrm{d}.&f(x)=\, ^{2}\log (x-1)\\ \textrm{e}.&f(x)=\, ^{2}\log (x+1) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{a}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\neq 0\: \: (\textrm{salah})\\ \textrm{b}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-2=2^{1}-2=0= 0\: \: (\textbf{benar})\\ \textrm{c}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-1=2^{1}-1=1\neq 0\: \: (\textrm{salah})\\ \textrm{d}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1-1)=\, ^{2}\log 0=\\ &\qquad\qquad\color{magenta}\textbf{tidak mungkin}\neq 0\: \: (\textrm{salah})\\ \textrm{e}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1+1)=\, ^{2}\log 2=1\neq 0\: \: (\textrm{salah})\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Solusi untuk persamaan}\: \: 3^{2x+1}=81^{x-2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&4,5\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{array}{|c|c|}\hline \begin{aligned}3^{2x+1}&=81^{x-2}\\ \left (3^{2x} \right ).3^{1}&=\left (3^{4} \right )^{x-2}\\ 3.\left (3^{2x} \right )&=3^{4x}.3^{-8}\\ 3.\left (3^{2x} \right )&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}\\ 0&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}-3\left ( 3^{2x} \right )\\ 0&=\left ( 3^{2x} \right )^{2}-3^{9}.\left ( 3^{2x} \right )\\ 0&=\left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right ) \end{aligned} &\begin{aligned}\textbf{atau}\qquad\qquad\qquad&\\ \left ( 3^{2x} \right )^{2}-27.\left ( 3^{2x} \right )&=0\\ \left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right )&=0\\ 3^{2x}=0\: \: \textrm{atau}\: \: 3^{2x}&=3^{9}\\ (\textrm{tm})\quad \textrm{atau}\: \: 3^{2x}&=3^{9}\\ 2^{x}&=9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{atau}\: \: x&=4\displaystyle \frac{1}{2} \end{aligned} \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ &\textrm{maka nilai dari}\: \: \left ( a+b-c \right )^{abc}=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&1000\\ \textrm{b}.&1\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&-1\\ \textrm{e}.&-1000 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\begin{aligned}\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )^{2}&=\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\times \left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\\ &=\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ &\qquad\qquad+\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ &=2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ &=2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ &=10+\sqrt{2^{2}.6}+\sqrt{2^{2}.10}+\sqrt{2^{2}.15}\\ &=10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}&=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ &\begin{cases} a &=2 \\ b & =3 \\ c &=5 \end{cases}.\qquad \textrm{sesuai dengan urutannya}\\ \textrm{Sehingga nilai}\qquad &\\ \left ( a+b-c \right )^{abc}&=\left ( 2+3-5 \right )^{2.3.5}\\ &=0^{30}\\ &=0 \end{aligned} \end{aligned} \end{array}$

Contoh Soal 1 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 1.&\textrm{Hasil dari}\: \: \displaystyle \frac{3^{4}.5^{3}.7^{2}}{27.125.49} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \textrm{d}.&9\\ \color{red}\textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\displaystyle \frac{3^{4}.5^{3}.7^{2}}{27.125.49}&=\displaystyle \frac{3^{4}.5^{3}.7^{2}}{3^{3}.5^{3}.7^{2}}\\ &=3^{4-3}.5^{3-3}.7^{2-2}\\ &=3^{1}.5^{0}.7^{0}\\ &=3.1.1\\ &=3 \end{aligned} \end{array}$

$\begin{array}{l}\\ 2.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{5a^{4}b^{2}}{a^{2}b^{-3}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab^{2}\\\\ \textrm{b}.&\displaystyle a^{2}b^{2}\\\\ \color{red}\textrm{c}.&\displaystyle 5a^{2}b^{5}\\\\ \textrm{d}.&5a^{4}b^{5}\\\\ \textrm{e}.&5a^{5}a^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \frac{5a^{4}b^{2}}{a^{2}b^{-3}}&=5a^{4-2}b^{2-(-3)}\\ &=5a^{2}b^{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Bentuk sederhana dari}\: \: \left ( \displaystyle \frac{ab^{2}}{a^{2}b^{3}} \right )^{4} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b}\\\\ \textrm{b}.&\displaystyle \frac{1}{ab}\\\\ \textrm{c}.&\displaystyle \frac{1}{a^{2}b^{2}}\\\\ \textrm{d}.&\displaystyle \frac{1}{ab^{2}}\\\\ \color{red}\textrm{e}.&\displaystyle \frac{1}{a^{4}b^{4}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{ab^{2}}{a^{2}b^{3}} \right )^{4}&=\left ( a^{1-2}b^{2-3} \right )^{4}\\ &=\left ( a^{-1}b^{-1} \right )^{4}\\ &=\left ( \displaystyle \frac{1}{ab} \right )^{4}\\ &=\displaystyle \frac{1}{a^{4}b^{4}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{2^{-\frac{1}{4}}\times 6^{\frac{2}{3}} }\: \: \textrm{adalah} ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6^{\frac{1}{4}}&&\textrm{d}.\quad \left ( \displaystyle \frac{2}{3} \right )^{\frac{3}{4}}\\ \color{red}\textrm{b}.\quad 6^{\frac{3}{4}}&\textrm{c}.\quad 6^{\frac{2}{3}}&\textrm{e}.\quad \left ( \displaystyle \frac{3}{2} \right )^{\frac{3}{4}}\end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \frac{3^{\frac{5}{6}}\times 12^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}&=\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4)^{\frac{7}{12}}}{(2\times 3)^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times \left ( 2^{2} \right )^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}\\ &=2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ &=2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ &=2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ &=(2\times 3)^{\frac{9}{12}}\\ &=6^{\frac{9}{12}}\\ &=6^{\frac{3}{4}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Bentuk sederhana dari}\: \: \left ( \displaystyle \frac{a^{\frac{1}{2}}b^{-3}}{a^{-1}b^{-\frac{3}{2}}} \right )^{\frac{3}{2}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a}{b}\\\\ \textrm{b}.&\displaystyle \frac{b}{a}\\\\ \textrm{c}.&\displaystyle ab\\\\ \textrm{d}.&\sqrt[a]{b}\\\\ \color{red}\textrm{e}.&\sqrt[4]{\left ( \displaystyle \frac{a}{b} \right )^{9}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{a^{\frac{1}{2}}b^{-3}}{a^{-1}b^{-\frac{3}{2}}} \right )^{\frac{3}{2}}&=\displaystyle \frac{a^{\frac{3}{4}}b^{\frac{-9}{2}}}{a^{-\frac{3}{2}}b^{-\frac{9}{4}}}\\ &=a^{\frac{3}{4}+\frac{3}{2}}b^{\frac{-9}{2}+\frac{9}{4}}\\ &=a^{\frac{3+6}{4}}b^{\frac{-18+9}{4}}\\ &=a^{\frac{9}{4}}b^{-\frac{9}{4}}\\ &=\left ( \displaystyle \frac{a}{b} \right )^{\frac{9}{4}}\\ &=\sqrt[4]{\left ( \displaystyle \frac{a}{b} \right )^{9}} \end{aligned} \end{array}$

Fungsi Eksponen

 $\LARGE\textrm{A. Bilangan Pangkat Positif}$

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:

$\color{blue}\LARGE a^{m}=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\color{magenta}\begin{aligned}\textrm{Bilangan}&:\\ \color{blue}a&\: \: \textrm{disebut basis atau bilangan pokok}\\ \color{blue}n&\: \: \textrm{disebut sebagai bilangan pangkat/eksponen} \end{aligned}$

$\LARGE\colorbox{yellow}{ Contoh Soal}$

$\color{blue}(1).\quad 3^{4}=3\times 3\times 3\times 3=81$

$\color{blue}(2).\quad 5^{4}=5\times 5\times 5\times 5=625$

$\color{blue}(3).\quad 2^{6}=2\times 2\times 2\times 2\times 2\times 2=64$

$\color{blue}(4).\quad 6^{7}=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$\color{blue}(5).\quad (-3)^{3}=(-3)\times (-3)\times (-3)=-27$

$\color{blue}(6).\quad (-2)^{6}=(-2)\times (-2)\times (-2)\times (-2)\times (-2)\times (-2)=64$

$\color{blue}(7).\quad \left ( \displaystyle \frac{1}{5} \right )^{3}=\left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )= \displaystyle \frac{1}{125}$

$\color{blue}(8).\quad \left ( -\displaystyle \frac{1}{2} \right )^{4}=\left ( -\displaystyle \frac{1}{2} \right )\times \left (- \displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )=- \displaystyle \frac{1}{32}$

$\LARGE\textrm{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{aligned}\\ 1.\quad&a^{m}.a^{n}=a^{m+n}\\ 2.\quad&a^{m}:a^{n}=a^{m-n}\\ 3.\quad&\left ( a^{m} \right )^{n}=a^{m.n},\: \: \textrm{syarat}\: \: a\neq 0\\ 4.\quad&\left ( ab \right )^{n}=a^{n}.b^{n}\\ 5.\quad&\left ( \frac{a}{b} \right )^{n}=\frac{a^{n}}{b^{n}},\: \: \textrm{syarat}\: \: b\neq 0 \end{aligned}$

Beberpa hal yang perlu diketahui juga, yaitu

$\color{blue}\begin{aligned}\\ 1.\quad&(a+b)^{2}=a^{2}+2ab+b^2\\ 2.\quad&(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\ 3.\quad&\left ( a+\frac{1}{a} \right )^{2}=a^{2}+2+\displaystyle \frac{1}{a^{2}},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\LARGE\colorbox{magenta}{ Contoh Soal}$

$\color{blue}\begin{aligned}\\ (1).\quad&2^{6} \times 2^{4} \times 2^{7} = 2^{6+4+7}=2^{17}\\ (2).\quad&2^{5} \times 3^{5} \times 7^{5} = \left ( 2 . 3 . 7 \right )^{5}=\left ( 42 \right )^{5}\\ (3).\quad&\displaystyle \frac{a^{3}.a^{7}.a^{6}}{a^{9}}=\displaystyle \frac{a^{3+7+6}}{a^{9}}=\frac{a^{16}}{a^{9}}=a^{16-9}=a^{7},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

$\begin{aligned}(4).\quad\displaystyle \frac{3^{7}.7^{3}.2}{\left ( 42 \right )^{3}}&=\frac{2^{1}.3^{7}.7^{3}}{\left ( 2.3.7 \right )^{3}}=\frac{2^{1}.3^{7}.7^{3}}{2^{3}.3^{3}.7^{3}}\\ &=2^{1-3}.3^{7-3}.7^{3-3}=2^{-2}.3^{4}.7^{0}\\ &=\frac{1}{2^{2}}.3^{4}.1=\frac{3^{4}}{2^{2}} \end{aligned}$

$\color{blue}\begin{aligned}(5).\quad\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}&=\displaystyle \frac{1.2^{2020}+2^{1}.2^{2020}+2^{2}.2^{2020}}{7}\\ &=\frac{\left ( 1+2+4 \right ).2^{2020}}{7}\\ &=\frac{7.2^{2020}}{7}\\ &=2^{2020} \end{aligned}$

$\color{blue}\begin{aligned}(6)\quad \displaystyle \frac{\left ( 2^{n+2} \right )^{2}-2^{2}.2^{2n}}{2^{n}.2^{n+2}}&=\displaystyle \frac{2^{2(n+2)}-2^{2}.2^{2n}}{2^{n}.2^{n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}.2^{2.2}-2^{2}.2^{2n}}{2^{n+n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}(2^{4}-2^{2})}{2^{2n}.2^{2}}\\ &=\displaystyle \frac{(2^{4}-2^{2})}{2^{2}}=\frac{16-4}{4}\\ &=\displaystyle \frac{12}{4}=3 \end{aligned}$

$\LARGE\textrm{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\color{magenta}\begin{aligned}&\\ 1.\quad&a^{ \frac{1}{n}}=\sqrt[n]{a}\\ 2.\quad&a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\\ 3.\quad&a^{\frac{1}{2}}=\sqrt[2]{a^{1}}=\sqrt{a} \end{aligned}$

Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\color{blue}\begin{aligned}&\\ 1.\quad&a\sqrt[n]{c}+b\sqrt[n]{c}=\left ( a+b \right )\sqrt[n]{c}\\ 2.\quad&a\sqrt[n]{c}-b\sqrt[n]{c}=\left ( a-b \right )\sqrt[n]{c}\\ 3.\quad&\sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.\quad&\sqrt[n]{a^{n}}=a\\ 5.\quad&a\sqrt[n]{c} x b\sqrt[n]{d} = ab\sqrt[n]{cd}\\ 6.\quad&\frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.\quad&\sqrt{\left ( a+b \right )+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.\quad&\sqrt{\left ( a+b \right )-2\sqrt{ab}}=\sqrt{a}-\sqrt{b} \end{aligned}$

Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\color{blue}\begin{aligned}1.\quad&\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left ( \sqrt{b^{2}} \right )}=\frac{a}{b}\sqrt{b}\\ 2.\quad&\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^{2}}}{\sqrt[3]{b^{2}}}=\frac{a\sqrt[3]{b^{2}}}{\left ( \sqrt[3]{b^{3}} \right )}=\frac{a}{b}\sqrt[3]{b^{2}}\\ 3.\quad&\displaystyle \frac{a}{\sqrt[5]{b^{3}}}=\displaystyle \frac{a}{\sqrt[5]{b^{3}}}\times \frac{\sqrt[5]{b^{2}}}{\sqrt[5]{b^{2}}}=\frac{a\sqrt[5]{b^{2}}}{\sqrt[5]{b^{5}}}=\frac{a}{b}\sqrt[5]{b^{2}} \end{aligned}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\color{blue}\begin{aligned}&\\ 1.\quad&\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c\left ( a-\sqrt{b} \right )}{a^{2}-b}\\ 2.\quad&\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c\left ( a+\sqrt{b} \right )}{a^{2}-b}\\ 3.\quad&\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c\left ( \sqrt{a}-\sqrt{b} \right )}{a-b}\\ \end{aligned}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^{2}}-\sqrt[3]{ab}+\sqrt[3]{b^{2}}$.

Contoh Soal 5 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 21.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \color{red}\textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\infty \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle  \frac{4}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle \frac{11}{4}\\ \color{red}\textrm{e}.&\displaystyle 60 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle  \frac{4}{x}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15}{u}.\tan 4u\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15\tan 4u}{u}\\ &=15\times 4\\ &=60 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \color{red}\textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=(ab)^{2} \end{aligned} \end{array}$

$\begin{array}{l}\\ 25.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \color{red}\textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{18}{x\sin \displaystyle \frac{3}{x}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -54 \\ \textrm{b}.&\displaystyle -6\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{6}\\ \color{red}\textrm{d}.&\displaystyle 6\\ \textrm{e}.&\displaystyle 54 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{18}{x\sin \displaystyle \frac{3}{x}}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18}{\displaystyle \frac{1}{u}\sin 3u}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18u}{\sin 3u}\\ &=\displaystyle \frac{18}{3}\\ &=6 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{4x\sin \displaystyle \frac{2}{x}}{2}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4 \\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{2}\\ \textrm{d}.&\displaystyle 2\\ \color{red}\textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{4x\sin \displaystyle \frac{2}{x}}{2}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{4}{u}.\sin 2u}{2}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{4\sin 2u}{2u}\\ &=\displaystyle \frac{4\times 2}{2}\\ &=4 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 29.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \color{red}\textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: x=2\: \: \textrm{dan}\: \: x=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \color{red}\textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=-3 \end{aligned} \end{array}$

Contoh Soal 4 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 16.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \color{red}\textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 19.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \color{red}\textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 20.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\color{red}\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=-2 \end{aligned} \end{array}$

Contoh Soal 3 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 11.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \color{red}\textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\color{magenta}\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{black}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=-\frac{5}{3} \end{aligned} \end{array}$

Contoh Soal 2 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

 $\begin{array}{ll}\\ 6.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 7.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }\\ &=\color{blue}\infty \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{magenta}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \color{red}\textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=2 \end{aligned} \end{array}$

$\color{magenta}\begin{aligned}\textrm{ada cara lain yang lebih sede}&\textrm{rhana, yaitu:}\\ .\qquad\: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ \textrm{Jika}\quad &\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}&-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ \textrm{Sehingga}\quad&\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=2 \end{aligned}$

$\begin{array}{ll}\\ 10.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\color{blue}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }=\infty \end{array}$

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$\color{magenta}\begin{aligned}\textrm{Sebagai}&\: \: \color{black}\textbf{CATATAN}\: \textrm{di sini}\\ \textrm{Sifat-sif}&\textrm{at bilangan tak hingga}\\ (1)\: \: &\infty +\infty =\infty \\ (2)\: \: &-\infty +(-\infty )=-\infty \\ (3)\: \: &\infty \times \infty =\infty\\ (4)\: \: &-\infty +(-\infty )=\infty \\ (5)\: \: &k.\infty =\infty ,\quad k\: \: \color{blue}\textrm{positif}\\ (6)\: \: &k.(-\infty )=-\infty,\quad k\: \: \color{blue}\textrm{positif} \\ (7)\: \: &k.\infty =-\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ (8)\: \: &k.(-\infty )=\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ \textrm{yang ha}&\textrm{rus dihindari}\\ (1)\: \: &\infty -\infty ,\quad \: \: \textrm{bentuk tak tentu}\\ (2)\: \: &\displaystyle \frac{\infty }{\infty },\: -\displaystyle \frac{\infty }{\infty },\: \: \textrm{dan}\: \: \frac{0}{0} \end{aligned}$

Contoh Soal 1 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 1.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\color{red}\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2020}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2021x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2020}{x^{2}}}{2-\displaystyle \frac{2021}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2020}{\infty ^{2}}}{2-\displaystyle \frac{2021}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2019}{\sqrt{9x^{2}-2020x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\color{red}\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2021}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2020x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2021}{x}}{-\sqrt{9-\displaystyle \frac{2020}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2021}{\infty }}{-\sqrt{9-\displaystyle \frac{2020}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\color{red}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \color{red}\textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e} \end{array}$

$\color{blue}\begin{aligned}.\qquad \: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: &\: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=-\displaystyle \frac{5}{3} \end{aligned}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=1 \end{aligned} \end{array}$

Limit di Ketakhinggaan (Kelas XII Matematika Peminatan)

 Untuk $n$ bilangan bulat positif, dan $k$ suatu konstanta, serta fungsi  $f$ dan $g$ adalah dua buah fungsi yang memiliki nilai limit saat  $x$  mendekati ketakhinggaan $\left (x\rightarrow \infty \right )$ , maka berlakulah sifat-sifat berikut:

$\begin{array}{ll}\\ 1.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{1}{x}=0\: \: \textbf{dan}\: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1}{x}=0\\ 2.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{1}{x^{n}}=0\: \: \textbf{dan}\: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1}{x^{n}}=0\\ 3.&\underset{x\rightarrow \infty }{\textrm{lim}}\: k=k\\ 4.&\underset{x\rightarrow \infty }{\textrm{lim}}\: k.f(x)=k.\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)\\ 5.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( f(x)+g(x) \right )=\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)+\underset{x\rightarrow \infty }{\textrm{lim}}\: g(x)\\ 6.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( f(x)-g(x) \right )=\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)-\underset{x\rightarrow \infty }{\textrm{lim}}\: g(x)\\ 7.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( f(x)\times g(x) \right )=\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)\times \underset{x\rightarrow \infty }{\textrm{lim}}\: g(x)\\ 8.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{f(x)}{g(x)}=\displaystyle \frac{\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)}{\underset{x\rightarrow \infty }{\textrm{lim}}\: g(x)}\\ 9.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left (f(x) \right )^{n}= \left [\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)) \right ]^{n}\\ 10.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \sqrt[n]{f(x)}=\sqrt[n]{\underset{x\rightarrow \infty}{\textrm{lim}}\: f(x)},\quad \textrm{dengan}\: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)\geq 0 \end{array}$

Contoh Soal 4 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 16.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-4}< \sqrt{2x+8} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-4<x\leq \displaystyle \frac{2}{3}\\ \textrm{b}.&-4<x<3\\ \color{red}\textrm{c}.&\displaystyle \frac{2}{3}\leq x< 3\\ \textrm{d}.&2<x\leq 4\\ \textrm{e}.&-4\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle \sqrt{6x-4}&< \sqrt{2x+8} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-4&<2x+8\\ 6x-2x&<8+4\\ 4x&<12\\ x&<3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-4&\geq 0\\ 6x&\geq 4\\ x&\geq \displaystyle \frac{2}{3}\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 2x+8&\geq 0\\ 2x&\geq -8\\ x&\geq -4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{x+3}> \sqrt{12-2x} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3<x\leq 6\\ \color{red}\textrm{b}.&-3<x\leq 6\\ \textrm{c}.&-6<x\leq 3\\ \textrm{d}.&-6<x\leq -3\\ \textrm{e}.&x<3\: \: \textrm{atau}\: \: x>6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \displaystyle \sqrt{x+3}&> \sqrt{12-2x} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x+3&>12-2x\\ x+2x&>12-3\\ 3x&>9\\ x&>3\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ x+3&\geq 0\\ x&\geq -3\\ 3.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 12-2x&\geq 0\\ 2x-12&\leq 0\\ 2x&\leq 12\\ x&\leq 6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&(\textbf{SBMPTN 2013 Mat Das})\\ &\textrm{Jika}\: \: 1<m<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x>-3\\ \textrm{b}.&x<-4\\ \color{red}\textrm{c}.&-4<x<0\\ \textrm{d}.&x<-4\: \: \textrm{atau}\: \: x>0\\ \textrm{e}.&x<-3\: \: \textrm{atau}\: \: x>-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{x^{2}+4x}{-x^{2}+3x-3m}>0\\ &\color{black}\textrm{dengan kondisi}\: \: 1<m<2\\ &\textrm{Perhatikanlah penyebutnya yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi penyebutnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena penyebutnya}\: :\: -x^{2}+3x-3m,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=3,\: \&\: c=-3m,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}3^{2}-4(-1)(-3m)=\color{black}9-12m\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}12<12m<24\\ &\Leftrightarrow \: \color{red}-12>-12m>-24\\ &\Leftrightarrow \: \color{red}9-12>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-3>\color{black}9-12m\color{red}>-13\\ &\Leftrightarrow \: \color{red}-13<\color{black}9-12m\color{red}<-3\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat penyebut berupa}\: \: -x^{2}+3x-3m\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{x(x+4)}{\underset{\color{red}definit\: negatif}{\underbrace{-x^{2}+3x-3m}}}>0\color{black}\Leftrightarrow \frac{x(x+4)}{-}>0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+4)<0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+4)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-4\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}-4<x<0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: 1<a<2\: ,\: \textrm{maka semua nilai}\\ &x\: \: \textrm{yang memenuhi pertidaksamaan}\\ &\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x<-3\: \: \textrm{atau}\: \: x>0\\ \textrm{b}.&x<-3\: \: \textrm{atau}\: \: x\geq -2\\ \textrm{c}.&x\leq -2\: \: \textrm{atau}\: \: x\geq 2\\ \textrm{d}.&-3<x<0\\ \textrm{e}.&-2\leq x<0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}1.\quad&\textrm{Diketahui bahwa}:\: \: \color{red}\displaystyle \frac{-x^{2}+2ax-6}{x^{2}+3x}\leq 0\\ &\textrm{untuk membedakan}\: \: a\: \: \textrm{pada persamaan}\\ &\textrm{kuadrat dengan}\: \: a\: \: \textrm{di atas, selanjutnya}\\ &\textrm{kita menuliskan}\: a\: \textrm{di atas dengan}:\: \color{red}m\\ &\color{black}\textrm{karena}\: \: 1<a<2\: \: \textrm{diubah}:1<m<2\\ &\textrm{Perhatikanlah pembilang yang}\\ &\textrm{mengandung bilangan}\: \: \color{red}m\: \: \color{blue}\textrm{yang terletak}\\ & \textrm{pada interval}\: :\: \color{red}1<m<2.\\ 2.\quad&\textrm{Kita cek kondisi pembilangnya dengan}\\ &\textrm{menentukan}\: Diskriminan(D)-\textrm{nya}\\ &\textrm{yaitu}:\\ &\color{red}ax^{2}+bx+c \color{black}\begin{cases} \color{black}a>0\: \&\: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit positif} \\ \color{red}a<0\: \& \: D=b^{2}-4ac<0\\ \qquad \Rightarrow \textrm{definit negatif} \end{cases}\\ &\textrm{Karena pebilangnya}\: :\: \color{red}-x^{2}+2mx-6,\\ &\textrm{dengan}\: \color{red}a=-1,\: b=2m,\: \&\: c=-6,\: \color{blue}\textrm{maka}\\ &\color{black}D=\color{red}(2m)^{2}-4(-1)(-6)=\color{black}4m^{2}-24\\ 3.\quad&\textrm{Karena nilai}\: \: \color{red}m\: \color{blue}\textrm{berada pada}\: \: \color{red}1<m<2\\ &\textrm{maka}\\ &\qquad \color{red}1<m<2\\ &\Leftrightarrow \: \color{red}1^{2}<m^{2}<2^{2}\color{blue}\Leftrightarrow \color{red}1<m^{2}<4\\ &\Leftrightarrow \: \color{red}4<4m^{2}<16\\ &\Leftrightarrow \: \color{red}4-24<\color{black}4m^{2}-24\color{red}<16-24\\ &\Leftrightarrow \: \color{red}-20<\color{black}4m^{2}-24\color{red}<-8\\ &\textrm{Ini berarti nilai}\: D\: \: \textrm{negatif, sehingga}\\ &\textrm{berakibat pembilangnya berupa}\: \: -x^{2}+2mx-6\\ &\color{black}\textrm{adalah wilayah}\: \: \color{red}definit\: negatif\\ 4.\quad&\textrm{Selanjutnya pemfaktoran pertidaksamaan}\\ &\bullet \: \color{black}\textrm{semula}\\ &\quad \displaystyle \frac{\overset{\color{red}definit\: negatif}{\overbrace{\color{blue}-x^{2}+2mx-6}}}{x(x+3)}\leq 0\color{black}\Leftrightarrow \frac{-}{x(x+3)}\leq 0\\ &\bullet \: \color{black}\textrm{akan berubah menjadi}\\ &\qquad \color{black}x(x+3)> 0\\ &\qquad\textrm{pembuat nol-nya adalah}:\: \color{red}x(x+3)=0\\ &\qquad\textrm{maka}\: \: \color{red}x=-3\: \: \color{blue}\textrm{atau}\: \color{red}x=0,\: \color{blue}\textrm{sehingga}\\ &\qquad \textrm{interval nilai}\: \color{red}x\color{blue}-\textrm{nya}\: : \color{red}x<-3\: \: \textrm{atau}\: \: x>0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&(\color{purple}\textrm{KSM 2019})\\ &\textrm{Banyaknya selesaian real dari persamaan}\\ &2x^{2}-7x+6=15\left \lfloor \displaystyle \frac{1}{x} \right \rfloor\left \lfloor x \right \rfloor\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&5\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{tidak ada}\\ & \end{array}$

$.\: \: \: \quad\begin{aligned}\textrm{Solu}&\textrm{si jawaban merujuk dari blog Pak Anang}\\ \textrm{And}&\textrm{a bisa mengunjunginya di blognya beliau} \end{aligned}$

$.\: \: \: \quad\begin{aligned}\textrm{di}& \end{aligned}$  sini (Soal dan Pembahasan KSM 2019 oleh Pak Anang)

Contoh Soal 3 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{ll}\\ 11.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x-5}\leq x \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1<x<0\\\\ \textrm{b}.&x<1\: \: \textrm{atau}\: \: x\geq 5\\\\ \color{red}\textrm{c}.&\displaystyle \frac{5}{6}\leq x\leq 1\: \: \textrm{atau}\: \: x\geq 5\\\\ \textrm{d}.&\displaystyle \frac{5}{6}\leq x< 1\: \: \textrm{atau}\: \: 5<x<6\\\\ \textrm{e}.&x\geq 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\sqrt{6x-5}&\leq x\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x-5&\leq x^{2}\\ -x^{2}+6x-5&\leq 0\\ x^{2}-6x+5&\geq 0\\ (x-1)(x-5)&\geq 0\\ x\leq 1\: \: \textrm{atau}&\: \: x\geq 5\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x-5&\geq 0\\ 6x&\geq 5\\ x&\geq \displaystyle \frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{6x+6}>6 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&x>7\\ \textrm{b}.&x\geq 7\\ \textrm{c}.&x<7\\ \textrm{d}.&x>1\\ \textrm{e}.&x\geq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\sqrt{6x+6}&>6\\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 6x+6&>36\\ x+1&>6\\ x&>7\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 6x+6&\geq 0\\ 6x&\geq 6\\ x&\geq \displaystyle \frac{6}{6}\\ x&\geq 1 \end{aligned} \end{array}$

$\begin{array}{l}\\ 13.&\textrm{Penyelesaian pertidaksamaan}\\ &x+2>\displaystyle \sqrt{10-x^{2}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\leq x\leq \sqrt{10}\\ \color{red}\textrm{b}.&1<x\leq \sqrt{10}\\ \textrm{c}.&-3<x\leq \sqrt{10}\\ \textrm{d}.&-\sqrt{10}\leq x\leq \sqrt{10}\\ \textrm{e}.&x< -3\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}x+2&>\displaystyle \sqrt{10-x^{2}} \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ x^{2}+4x+4&>10-x^{2}\\ 2x^{2}+4x&+4-10>0\\ 2x^{2}+4x-&6>0\\ x^{2}+2x-&3>0\\ (x+3)&(x-1)>0\\ x<-3&\: \: \textrm{atau}\: \: x>1\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 10-x^{2}&\geq 0\\ x^{2}-10&\leq 0\\ (x-\sqrt{10})&(x+\sqrt{10})\leq 0\\ -\sqrt{10}\leq x&\leq \sqrt{10} \end{aligned} \end{array}$

$\begin{array}{l}\\ 14.&\textrm{Penyelesaian pertidaksamaan}\\ &\displaystyle \sqrt{3x+7}\geq x-1 \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-1<x<6\\ \textrm{b}.&-1\leq x<6\\ \textrm{c}.&x\geq -\displaystyle \frac{7}{3}\\ \color{red}\textrm{d}.&-\displaystyle \frac{7}{3}\leq x\leq 6\\ \textrm{e}.&-\displaystyle \frac{7}{3}\leq x\leq 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sqrt{3x+7}&\geq x-1 \\ 1.\quad\textrm{Kuadrat}&\textrm{kan}\\ 3x+7&\geq x^{2}-2x+1\\ -x^{2}+3x&+2x+7-1\geq 0\\ -x^{2}+5x+&6\geq 0\\ x^{2}-5x-&6\leq 0\\ (x+1)&(x-6)\leq 0\\ -1\leq x&\leq 6\\ 2.\quad \textrm{Di bawa}&\textrm{h tanda akar}\: \: \geq 0\\ 3x+7&\geq 0\\ 3x&\geq -7\\ x&\geq -\displaystyle \frac{7}{3} \end{aligned} \end{array}$

 $\begin{array}{ll}\\ 15.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x\geq -5\\ \textrm{b}.&x<-13\: \: \textrm{atau}\: \: x\geq -4\\ \textrm{c}.&x<13\\ \color{red}\textrm{d}.&4\leq x< 13\\ \textrm{e}.&-5\leq x\leq 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sqrt{2x-8}&<\sqrt{x+5}\\ (1)\: \: \, \textrm{kuadratkan}&\\ 2x-8&<x+5\\ x&<13\\ (2)\quad 2x-8\geq 0&\\ x&\geq 4\\ (3)\: \: \quad x+5\geq 0&\\ x&\geq 5\\ \textrm{perhatikan}&\textrm{lah garis bilangannya berikut}\\ 1 \qquad&\color{black}\begin{array}{ccccccc|ccccc}\\ &&&&&&&&&&\\\cline{1-7} &&&&&&&&&&\textrm{X}\\\hline &&&&&&13&&&& \end{array}\\\\ 2\qquad&\color{black}\begin{array}{ccc|ccccccccc}\\ &&&&&&&&&&\\\cline{4-11} &&&&&&&&&&\textrm{X}\\\hline &&&4&&&&&&& \end{array}\\\\ 3\qquad&\color{black}\begin{array}{ccccc|ccccccc}\\ &&&&&&&&&&\\\cline{6-11} &&&&&&&&&&\textrm{X}\\\hline &&&&&5&&&&& \end{array} \\ \end{aligned} \end{array}$