CONTOH 10-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{aligned}46.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}=\:....\\ &\text{a}.\quad \displaystyle \frac{2}{3}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 3\\ &\text{b}.\quad 1\qquad\qquad \text{c}.\quad 2\qquad\qquad \text{e}.\quad 4\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Ingat bentuk}\\ & \underset{x\rightarrow \infty }{\textrm{Lim}}\: \sqrt{\displaystyle ax^{\displaystyle 2}+bx+c}-\sqrt{\displaystyle ax^{\displaystyle 2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga soal untuk di atas}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-(1+x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1}}\\ &=\displaystyle \frac{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1} \right)}{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1} \right)}\\ &=\displaystyle \frac{\left( \displaystyle \frac{-3-(-1)}{2\sqrt{1}} \right)}{\left( \displaystyle \frac{1-2}{2\sqrt{1}} \right)}=\displaystyle \frac{-2}{-1}=2\\\end{aligned}$

$\begin{aligned}47.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\:....\\ &\text{a}.\quad \displaystyle 1\:\:\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 9\\ &\text{b}.\quad 3\qquad\qquad \text{c}.\quad 4\qquad\qquad \text{e}.\quad 16\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x}.2+3^{\displaystyle x}.3}{\displaystyle \frac{2^{\displaystyle x}}{2}+\displaystyle \frac{3^{\displaystyle x}}{3}}\times \frac{\displaystyle \frac{1}{3^{\displaystyle x}}}{\displaystyle \frac{1}{3^{\displaystyle x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}\\ & \textrm{Perhatikan bahwa saat}\: x\longrightarrow \infty \:,\: \textrm{maka}\:\: \left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}\longrightarrow 0\\ &\textrm{Sehingga}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}= \displaystyle \frac{2.\left( 0 \right)+3}{\displaystyle \frac{1}{2}.\left( 0 \right)+\displaystyle \frac{1}{3}}=\displaystyle \frac{3}{\displaystyle \frac{1}{3}}=9 \end{aligned}$.

$\begin{aligned}48.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=\:....\\\\ &\textrm{dengan}\:\left\lfloor x \right\rfloor= \textrm{bilangan bulat terbesar yang }\\ &\textrm{kurang dari atau sama dengan}\:\:\: x\\\\ &\text{a}.\quad \displaystyle \frac{1}{2}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 1\\ &\text{b}.\quad 2\qquad\qquad \text{c}.\quad 0\qquad\qquad \text{e}.\quad \textrm{tidak ada}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Berdasarkan sifat fungsi tangga, untuk setiap bilangan}\\ &\textrm{riil}\:\:\: x\:\:\: \textrm{berlaku}:\\ & x-1<\left\lfloor x \right\rfloor\le  x\\ &\text{Untuk}\quad x>0,\:\: \textrm{bagilah seluruh ruas dengan}\:\:\: x\\ &\displaystyle \frac{x-1}{x}<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le \displaystyle \frac{x}{x}\Leftrightarrow \left( 1-\displaystyle \frac{1}{x} \right)<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le 1\\ &\textrm{Selanjutnya kita hitung nilai limit batas kiri dan kanan}\\ &\text{ketika}:x\longrightarrow \infty \:(\textrm{ingat ini bukan limit kiri dan kanan})\\ &\circ \quad \textrm{Batas kiri}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( 1-\displaystyle \frac{1}{x} \right)=1-\displaystyle \frac{1}{\infty }=1-0=1\\ &\circ \quad \textrm{Batas kanan}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: 1=1\\ &\textrm{Berdasarkan teorema apit (Squeeze Theorem), karena}\\ &\textrm{batas kiri sama dengan batas kanan, maka nilai}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=1 \end{aligned}  \end{aligned}$.

CONTOH 9-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{ll}\\ 41.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{18}{x\sin \displaystyle \frac{3}{x}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -54 \\ \textrm{b}.&\displaystyle -6\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{6}\\ \textrm{d}.&\displaystyle 6\\ \textrm{e}.&\displaystyle 54 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{18}{x\sin \displaystyle \frac{3}{x}}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18}{\displaystyle \frac{1}{u}\sin 3u}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18u}{\sin 3u}\\ &=\displaystyle \frac{18}{3}\\ &=6 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 42.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{4x\sin \displaystyle \frac{2}{x}}{2}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4 \\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{2}\\ \textrm{d}.&\displaystyle 2\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle  \frac{4x\sin \displaystyle \frac{2}{x}}{2}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{4}{u}.\sin 2u}{2}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{4\sin 2u}{2u}\\ &=\displaystyle \frac{4\times 2}{2}\\ &=4 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 44.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: x=2\: \: \textrm{dan}\: \: x=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=-3 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Astuti, A.N., Miyanto, Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA
2. Kartini, Suprpto, Subandi, Setiyadi, U. 2005.Matematika Program Studi Ilmu ALam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Noormandiri. 2017. Matematika Kelompok Peminatan Matematika dan Ilmu-Ilmu ALam untuk SMA/MA Kelas XII. Jakarta: ERLANGGA
4. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
5. Tim. 2020. Modul Matematika (Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII). Tangerang: RAHMA GEMILANG.

CONTOH 8-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\infty \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle  \frac{4}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle \frac{11}{4}\\ \textrm{e}.&\displaystyle 60 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle  \frac{4}{x}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15}{u}.\tan 4u\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15\tan 4u}{u}\\ &=15\times 4\\ &=60 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=(ab)^{2} \end{aligned} \end{array}$

$\begin{array}{l}\\ 40.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=0 \end{aligned} \end{array}$

CONTOH 7-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{ll}\\ 31.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{b}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 34.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 35.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=-2 \end{aligned} \end{array}$

CONTOH 6-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{ll}\\ 26.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{black}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=-\frac{5}{3} \end{aligned} \end{array}$

CONTOH 5-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{ll}\\ 21.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 22.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\sqrt{\infty }+\sqrt{\infty }\\ &=\infty \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=2 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\sqrt{\infty }+\sqrt{\infty }=\infty \end{array}$

☝☝☝☝☝☝☝☝☝☝☝☝

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

CONTOH 4-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

 $\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2026}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2027x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2026}{x^{2}}}{2-\displaystyle \frac{2027}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2026}{\infty ^{2}}}{2-\displaystyle \frac{2027}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2027}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2028x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2027}{x}}{-\sqrt{9-\displaystyle \frac{2028}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2027}{\infty }}{-\sqrt{9-\displaystyle \frac{2028}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e} \end{array}$

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

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=1 \end{aligned} \end{array}$

CONTOH 3-LIMIT FUNGSI

 $\begin{aligned}11.\quad&\textrm{Perhatikan gambar grafik fungsi}\:\: y=f(x)\: \: \textrm{berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: a\in \left\{ 0,2,4 \right\}\:\: \textrm{dengan}\:\: \underset{x\rightarrow a }{\textrm{Lim}}\: f(x)\:\: \textrm{ada, maka}\\ &\textrm{untuk nilai semua}\:\: a\:\: \textrm{yang memenuhi adalah}\: ....\\ &\textrm{a}.\quad a=0\\ &\textrm{b}.\quad a=2\\ &\textrm{c}.\quad a=0\:\: \textrm{dan}\:\: a=2\\ &\textrm{d}.\quad a=0\:\: \textrm{dan}\:\: a=4\\ &\textrm{e}.\quad a=0\:,\: a=2\:,\: \textrm{dan}\:\: a=4\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas sesuai ilustrasi gambar, bahwa}\\ &\bullet \quad\underset{\displaystyle x\rightarrow  0^{\displaystyle  -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow  0^{\displaystyle  +} }{\textrm{Lim}\qquad}\: f(x)=6\\ &\textrm{dan}\\ &\bullet \quad \underset{\displaystyle x\rightarrow  2^{\displaystyle  -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow  2^{\displaystyle  +} }{\textrm{Lim}\qquad}\: f(x)=2\\ &\textrm{serta nilai}\\ &\bullet \quad \underset{\displaystyle x\rightarrow  4^{\displaystyle  -} }{\textrm{Lim}\qquad}\: f(x)\neq \underset{\displaystyle x\rightarrow  4^{\displaystyle  +} }{\textrm{Lim}\qquad}\: f(x)  \end{aligned}$.

$\begin{array}{l}\\ 12.&\textrm{Diketahui}\\ &f(x)=\begin{cases} 3x^{\displaystyle 2}+8x+1 & ,&\textrm{untuk}\quad x<2 \\ ax+3&,&\textrm{untuk}\quad x\ge 2 \end{cases}\\ &\textrm{Besar}\quad  a\quad \textrm{yang memenuhi agar nilai}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: f(x)\:\: \textrm{ada}\quad\textrm{adalah}\: ....\\  &\begin{array}{llllllll} \textrm{a}.&-13\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 13\\ \textrm{b}.&-2\qquad\qquad \textrm{c}.\quad 2\:\quad\quad\qquad \textrm{e}.\quad 26\\  \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Suatu fungsi}\quad f(x)\quad \textrm{memiliki nilai limit}\:\:  \underset{x\rightarrow 2}{\textrm{Lim}}\: f(x)\\&\textrm{jika dan hanya jika nilai limit kiri sama dengan}\\ &\textrm{limit kanan pada}\quad x=2,\quad \textrm{yaitu}: \\ &\begin{aligned}&\underset{.\quad\displaystyle x\rightarrow 2^{\displaystyle -}}{\textrm{Lim}} f(x)\quad=\underset{.\quad\displaystyle x\rightarrow  2^{\displaystyle +}}{\textrm{Lim}}\: f(x)\\ &\Leftrightarrow 3(2)^{\displaystyle 2}+8(2)+1=a(2)+3\Leftrightarrow 29=2a+3\\ &\Leftrightarrow 2a=29-3\Leftrightarrow 2a=26\Leftrightarrow a=13\\  \end{aligned} \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Diketahui}\quad f(x)\quad \text{adalah fungsi dari polinom berderajat}\\ &3\quad \textrm{yang memenuhi kondisi}\quad \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x+1}=6\quad \textrm{dan}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x-2}=30\:,\: \textrm{maka fungsi}\quad f(x)=\: ....\\  &\begin{array}{llllllll} \textrm{a}.&4x^{\displaystyle 3}+2x^{\displaystyle 2}-10x-4\quad\quad\textrm{d}.\quad 4x^{\displaystyle 3}-2x^{\displaystyle 2}+10x-4\\ \textrm{b}.&4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x+4\quad\quad \textrm{e}.\quad 4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x-4\\  \textrm{c}.&4x^{\displaystyle 3}-2x^{\displaystyle 2}+10x+4\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textbf{Analisis polinom (suku banyak)}\\ &\circ \quad (x+1)\quad \textrm{dan}\quad (x-2)\quad \textrm{adalah faktor dari}\quad f(x)\\&\circ \quad f(x)=(x+1)(x-2)(ax+b)\quad \textrm{dikalikan dengan}\\ &\:\:\quad (ax+b)\quad \textrm{agar menjadi polinom berderjat 3}\\ &\textbf{Kondisi limit}\\&\circ \quad \textrm{limit pertama}\\  &\:\:\quad \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x+1}=6\Leftrightarrow \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{(x+1)(x-2)(ax+b)}{x+1}=6\\ &\:\:\quad \Leftrightarrow  \underset{x\rightarrow -1 }{\textrm{Lim}}\: (x-2)(ax+b)=6 \\ &\:\:\quad \textrm{selanjutnya kita substitusikan}\quad x=-1,\quad \textrm{yaitu}:\\ &\:\:\quad (-1-2)(a(-1)+b)=6\Leftrightarrow -a+b=-2\: .......(1)\\&\circ \quad \textrm{limit kedua}\\  &\:\:\quad \underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x-2}=30\Leftrightarrow \underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{(x+1)(x-2)(ax+b)}{x-2}=30\\ &\:\:\quad \Leftrightarrow  \underset{x\rightarrow 2 }{\textrm{Lim}}\: (x+1)(ax+b)=30 \\ &\:\:\quad \textrm{selanjutnya kita substitusikan}\quad x=2,\quad \textrm{yaitu}:\\ &\:\:\quad (2+1)(a(2)+b)=30\Leftrightarrow 2a+b=10\: .......(2)\\ &\textbf{Menentukan nilai a dan b}\\ &\textrm{dengan eliminasi akan kita peroleh}\\ &\begin{array}{rl} -a+b=-2&\\ 2a+b=10&-\\\hline -3a=-12&\Leftrightarrow a=4\: \Rightarrow b=2\end{array}\\ &\textbf{Menyusun}\quad f(x)\\&f(x)=(x+1)(x-2)(4x+2)=4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x-4\\ \end{array}$.

$\begin{array}{l}\\ 14.&\textrm{Sebuah virus menyebar luas di suatu daerah dengan }\\ &\textrm{sangat cepat dan luas daerah sebarannya dirumuskan}\\ &A(t)=\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}\quad \textrm{dengan}\quad t\:\: \textrm{adalah hari penyebaran}\\ &\textrm{virus. Laju penyebaran virus  pada hari ke}-t\\ &\textrm{diformulasikan dengan rumus}\quad \underset{t\rightarrow t_{1} }{\textrm{Lim}}\: \displaystyle \frac{A(t)-A\left( t_{1} \right)}{t-t_{1}}.\:\\ &\textrm{Laju sebaran virus pada hari ke-7 sejak wabah}\\&\textrm{menyebar adalah}\quad(\textrm{luas dalam}\:\: m^{\displaystyle 2})\: ....\: \displaystyle \frac{m^{\displaystyle 2}}{\textrm{hari}}\\  &\begin{array}{llllllll} \textrm{a}.&0,8\:\:\qquad\qquad\qquad\qquad\quad\qquad \textrm{d}.\quad 2,4\\ \textrm{b}.&1,2\qquad\qquad \textrm{c}.\quad 1,6\:\quad\quad\qquad \textrm{e}.\quad 2,8\\  \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Diketahui fungsi luas}\quad A(t)=\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}\:\: \textrm{dan rumus laju}\\ & \textrm{penyebaran virus}\quad V=\underset{t\rightarrow t_{1} }{\textrm{Lim}}\: \displaystyle \frac{A(t)-A\left( t_{1} \right)}{t-t_{1}}\: ,\: \textrm{maka hari} \\&\textrm{ke-7 laju sebaran virusnya}\:\: \left( t_{1}=7 \right)\:\: \textrm{adalah}:\\ &V=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}-\displaystyle \frac{1}{36}\pi (7)^{\displaystyle 2}}{t-7}=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\pi(t^{\displaystyle 2}-49)}{t-7}\\&\quad =\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(t+7)(t-7)}{t-7}=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(t+7)\\ &\quad =\displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(7+7)=\displaystyle \frac{44}{36}=1,\overline{222}\approx 1,2 \end{array}$.

$\begin{array}{l}\\ 15.&\textrm{Dalam sebuah riset terkait peningkatan penggunaan}\\ &\textrm{data internet pada perangkat seluler, pertumbuhan}\\ &\textrm{penggunaan data (dalam gigabita) dimodelkanan}\\ &\textrm{oleh fungsi}\:\:f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{dengan}\:\:x\:\: \textrm{adalah}\\ &\textrm{waktu dalam bulan. Jika kecenderungan peningkatan}\\ &\textrm{waktu yang dihasilkan oleh fungsi}\quad  f(x)\quad \textrm{terus}\\&\textrm{berlanjut, perkiraan penggunaan data mendekati}\\ &\textrm{tak hingga adalah}\: ....\: \textrm{gigabita}\\  &\begin{array}{llllllll} \textrm{a}.&300\:\:\qquad\qquad\qquad\qquad\quad\qquad \textrm{d}.\quad 100\\ \textrm{b}.&200\qquad\qquad \textrm{c}.\quad 150\:\quad\quad\qquad \textrm{e}.\quad 50\\  \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Diketahui fungsi }\quad f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{untuk}:x\to \infty \\ & \textrm{maka}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: f(x)=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\times \displaystyle \frac{\displaystyle \frac{1}{x^{\displaystyle 2}}}{\displaystyle \frac{1}{x^{\displaystyle 2}}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100+\displaystyle \frac{200}{x}}{2+\displaystyle \frac{100}{x^{\displaystyle 2}}}\\&=\displaystyle \frac{100+\displaystyle \frac{200}{\infty }}{2+\displaystyle \frac{100}{\infty ^{\displaystyle 2}}}=\displaystyle \frac{100+0}{2+0}=50\\ \end{array}$.

CONTOH 2-LIMIT FUNGSI

 $\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &= 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &= \displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad  5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &= 5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}= \displaystyle \frac{2}{14} \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai}\quad  \underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}\quad \textrm{adalah}\: ....\\  &\begin{array}{llllllll} \textrm{a}.&6\sqrt{5}\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 36\sqrt{5}\\ \textrm{b}.&12\sqrt{5}\qquad \textrm{c}.\quad 24\sqrt{5}\:\quad\qquad \textrm{e}.\quad 48\sqrt{5}\\  \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}=\displaystyle \frac{10^{\displaystyle 2}-14\times 10+40}{\sqrt{20}-\sqrt{10+10}}=\displaystyle \frac{0}{0}\\ &\textrm{bentuk di atas harus dihindari.  Selanjutnya}\\ &\textrm{gunakan opsi mengalikan penyebut dengan sekawan},\\ &\textrm{yaitu:}\\ &\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}\times \frac{\left( \sqrt{2x}+\sqrt{x+10} \right)}{\left( \sqrt{2x}+\sqrt{x+10} \right)}\\ &= \underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{\left( x^{\displaystyle 2}-14x+40 \right)\left( \sqrt{2x}+\sqrt{x+10} \right)}{2x-(x+10)}\\ &=\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{(x-10)(x-4)\left( \sqrt{2x}+\sqrt{x+10} \right)}{x-10}\\ &=\underset{x\rightarrow 10}{\textrm{Lim}}\: (x-4)\left( \sqrt{2x}+\sqrt{x+10} \right)\\ &=(10-4)\left( \sqrt{20}+\sqrt{20} \right)\\ &=(6)\left( 2\sqrt{20} \right)=(6)(2)\left( 2\sqrt{5} \right)=24\sqrt{5}\end{aligned} \end{array}$.

CONTOH 1-LIMIT FUNGSI

 $\begin{array}{ll}\\ 1&\textrm{Nilai}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle \frac{1}{4}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{4}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&= \left ( \displaystyle \frac{6-2}{2^{2}-4}-\frac{1}{2-2} \right )\\ &=\left ( \displaystyle \frac{4}{0}-\frac{1}{0} \right )= \infty -\infty \\ &\: \: \textrm{hal ini tidak diperkenankan}\\ \textrm{Sehingga},\, \qquad\qquad\qquad &\\ \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{(x+2)}{(x-2)(x+2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{x+2}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{4-2x}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2(x-2)}{(x+2)(x-2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2}{x+2} \right )\\ &=\displaystyle -\frac{2}{(2+2)}\\ &= -\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 2&&\textrm{d}.\quad \displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle 2\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&= \left ( \displaystyle \frac{4-4}{2^{4}-4} \right )\\ &= \displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\: \: \: \: \quad&\\ \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{\sqrt{x}\left ( 2-\sqrt{x} \right )} \\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{-\sqrt{x}\left ( \sqrt{x}-2 \right )}\\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: -\displaystyle \frac{\left ( \sqrt{x}+2 \right )}{\sqrt{x}}\\ &=-\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}\\ &=-\displaystyle \frac{2+2}{2}\\ &= -2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 1\quad &\textrm{e}.\quad \displaystyle 4\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 1}{\textrm{Lim}}\:& \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &= \displaystyle \frac{\left ( 2-3+1 \right )\left ( 1-1 \right )}{\left ( 1-1 \right )^{2}} \\ &=\displaystyle \frac{0\times 0}{0^{2}}\\ &= \displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehi}&\textrm{ngga},\\ \underset{x\rightarrow 1}{\textrm{Lim}}\: &\: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \left ( 2\sqrt{x}-1 \right )\times \left ( \sqrt{x}-1 \right ) \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-1 \right )}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \left ( 2\sqrt{x}-1 \right )\\ &=2.1-1\\ &=2-1\\ &=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{14}\sqrt{7}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{7}\sqrt{7}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{14}\sqrt{7}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 3}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times \displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x+4 \right )-\left ( 2x+1 \right )}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: -\displaystyle \frac{-x+3}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{-1}{\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=-\displaystyle \frac{1}{\left ( \sqrt{7}+\sqrt{7} \right )}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\times \displaystyle \frac{\sqrt{7}}{\sqrt{7}}\\ &= -\displaystyle \frac{1}{14}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\: \: \textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad \displaystyle -2\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad -1\quad &\textrm{e}.\quad \displaystyle -3\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: &\: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6\\ &\textrm{dengan bantuan limit kanan }\\ &\textrm{yaitu}\: \: x=2+h\: \Rightarrow \: h\rightarrow 0\\ \underset{h\rightarrow 0}{\textrm{Lim}}\: &\: \displaystyle \frac{a\left ( 2+h \right )-2a}{\sqrt{2\left ( 2+h \right )}-\left ( 2+h \right )}=6\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-\left ( 2+h \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah}{\sqrt{4+2h}-\left ( 2+h \right )}\times \displaystyle \frac{\left ( \sqrt{4+2h}+\left ( 2+h \right ) \right )}{\left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{4+2h-\left ( 4+4h+h^{2} \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2h-h^{2}}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2-h}\\ 6&=\displaystyle \frac{a\times \left (\sqrt{4+0} +\left ( 2+0 \right ) \right )}{-2-0}\\ 6&=\displaystyle \frac{a\left ( \sqrt{4}+2 \right )}{-2}\\ \displaystyle \frac{a(4)}{-2}&=6\\ a(-2)&=6\\ a&= -3 \end{aligned} \end{array}$

LIMIT FUNGSI

 Kumpulan materi yang sudah ada

• materi limit aljabar
• contoh soal 1, contoh 2, 

• limit fungsi trigonometri (materi lama-perkenalan dengan limit fungsi aljabar), lanjutan materi lama, lanjutan rumus dasar limit trigonometri, 
• materi pendahuluan limit, menentukan nilai limit fungsi trigonometri, pembuktian teorema apit (identitas limit fungsi trigonometri), limit fungsi trigonometri di tak hingga,
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,
• materi limit diketakhinggan.
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,

ALHAMDULILLAH-HASIL OSKANU-MA FUTUHIYAH JEKETRO GUBUG GROBOGAN-SEPTEMBER 2026

 

CONTOH SOAL 13 LINGKARAN (LINGKARAN DAN SEGITIGA)

 $\begin{aligned}61.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Titik O adalah pusat lingkaranyang berjari-jari}\:\: r\\ &\textrm{Jika panjang ruas garis}\: ED=r,\:\: \textrm{maka rasio}\\ &\angle CED\:\: \textrm{terhadap}\:\: \angle AOB\:\: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 1:3\\ &\textrm{b}.\quad 1:2\\ &\textrm{c}.\quad 2:3\\ &\textrm{d}.\quad 2:5\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Bukti ada pada soal nomor 62}\\  \end{aligned}$.

$\begin{aligned}62.\quad&\textrm{Buktikan jawaban pada soal nomor 61 di atas}\\\\ &\textrm{Bukti}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Misalkan besar}\:\: \angle CED=x^{\displaystyle 0}\:\: \textrm{dan panjang ruas garis}\\ &ED=OA=OB=OD=r,\:\: \textrm{akibatnya}\:\: \bigtriangleup ODE\\ &\textrm{dan}\:\: \bigtriangleup OBD\:\: \textrm{adalah sama-sama segitiga sama kaki}\\ &\textrm{tetapi beda besar sudutnya. Jika untuk}\:\: \angle CED=x^{\displaystyle 0},\\ &\text{maka}\:\: \angle ODB=2x^{\displaystyle 0},\:\:  \angle CED=\angle EOD=x^{\displaystyle 0}.\\ &\textrm{Akibat lanjutannya juga, yaitu}:\: \angle AOB=3x^{\displaystyle 0},\\ &\text{karena}\:\: \angle ODB=\angle OBD=2x^{\displaystyle 0}.\:\: \textrm{Besar }\: \angle AOB=3x^{\displaystyle 0},\\ &\textrm{karena akibat dari sudut luar}\:\: \bigtriangleup EOB\\ &\textrm{Jadi, rasio}\:\: \angle DEC:\angle AOB=x^{\displaystyle 0}:3x^{\displaystyle 0}=1:3\quad (\textbf{terbukti})   \end{aligned}$.

$\begin{aligned}63.\quad&(\textbf{LM UGM ke-26 Th 2015 Tk.SMA})\\ &\textrm{Perhatikan gambar berikut}\end{aligned}$

$\begin{aligned}\quad\qquad&\textrm{Diketahui lingkaran tersebut berpusat di O dengan}\\ &\textrm{dua buah garis singgung lingkaran masing-masing di}\\ &\textrm{titik A dan B dan dua garis singgung tersebut berpo-}\\ &\textrm{tongan di titik P. Garis QR juga menyinggung ling-}\\ &\textrm{karan di S. Jika diameter lingkaran = 14, serta pan-}\\ &\textrm{jang}\:\:OP=25,\:\: \textrm{keliling}\:\: \bigtriangleup PQR\:\: \:  \textrm{adalah}\: ....\\  &\textrm{a}.\quad 24\\ &\textrm{b}.\quad 48\\ &\textrm{c}.\quad 72\\ &\textrm{d}.\quad 96\\ &\textrm{e}.\quad 120 \\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}:PA=PB=24\:\: \textrm{ingat untuk}\\ &\textrm{tripel Pythagoras}:(7,24,25),\: \textrm{karena},\: r=7.\\ &\textrm{Perhatikan pula bahwa}: QS=QA\: \textrm{dan}\: RB=RS\\ &\textrm{akibat dari garis singgung lingkaran. Sehingga}\\ &\textrm{keliling}\:\: \bigtriangleup PQR=PQ+QS+SR+RP\\ &\:\,\qquad\qquad\qquad\quad=PQ+QA+BR+RP\\ &\:\,\qquad\qquad\qquad\quad=PA+BP\\ &\:\,\qquad\qquad\qquad\quad=24+24\\ &\:\,\qquad\qquad\qquad\quad=48  \end{aligned}$.