Nama: - | ID: -
0
0
0
$\begin{aligned}3.\quad &(\textbf{LM UGM ke-19 Th.2007 Tk. SMA})\\ &\textrm{Diketahui barisan geometri dengan suku pertama}\quad a\\ &\textrm{dan rasio}\quad r.\:\: \textrm{Untuk sebarang}\quad n\quad \textrm{genap didefinisikan}\\ &\textrm{jumlahan}\quad S_{\displaystyle n}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+...+U_{\displaystyle n}\quad \textrm{dan}\\ &\widehat{S}_{\displaystyle n}=U_{\displaystyle 1}-U_{\displaystyle 2}+U_{\displaystyle 3}-U_{\displaystyle 4}+...+U_{\displaystyle n-1}-U_{\displaystyle n}.\quad \textrm{Nilai}\\ &r\quad \textrm{yang mungkin agar}\quad \displaystyle \frac{\widehat{S}_{\displaystyle n}}{S_{\displaystyle n}}>3?\\ &\text{a}.\quad -1< r<0\\ &\text{b}.\quad -1< r<\displaystyle -\frac{1}{2}\\ &\text{c}.\quad -1< r<\displaystyle -\frac{1}{3}\\ &\text{d}.\quad \displaystyle -\frac{1}{3}< r<0\\ &\text{e}.\quad \textrm{tergantung oleh nilai}\quad a\\ \end{aligned}$
.$\begin{aligned}4.\quad&(\textbf{OLIMPIADE SAINS PORSEMA Th.2012})\\ &\textrm{Jumlah 50 suku pertama dari deret berikut}\\ &\log 5+\log 55+ \log 605+\log 6655\:+\:...\quad \textrm{adalah}\: ....\\ &\text{a}.\quad \log \left( 55^{\displaystyle 1155} \right)\qquad\qquad\qquad\text{d}.\quad \log \left( 275^{\displaystyle 1150} \right)\\ &\text{b}.\quad \log \left( 5^{\displaystyle 25}.11^{\displaystyle 1225} \right)\qquad\quad\quad\text{e}.\quad 1150\log 5\\ &\text{c}.\quad \log \left( 25^{\displaystyle 25}.11^{\displaystyle 1225} \right)\\\end{aligned}$
$\begin{aligned}1.\quad&(\textbf{LM UGM ke-25 Th 2014 Tk.SMA})\\ &\textrm{Diberikan dua buah barisan aritmetika}\quad 1,4,7,...\quad \textrm{dan}\\ &2,7,12,...\:. \: \textrm{Jika}\:\:S\:\: \textrm{merupakan himpunan yang terdiri}\\ &\textrm{dari gabungan 2014 suku pertama kedua barisan tersebut},\\ &\textrm{Banyak anggota himpuan}\:\: S\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad 3625\qquad\qquad\qquad\qquad \text{d}.\quad 4015\\ &\text{b}.\quad 3875\qquad \text{c}.\quad 4014\qquad \text{e}.\quad 4028\\ \end{aligned}$
.$\begin{aligned}2.\quad&(\textbf{LM UGM Ke-25 Th 2014 Tk.SMA})\\ &\textrm{Hitunglah nilai dari}\\ &5-\displaystyle \frac{10}{3}+\frac{20}{9}-\frac{40}{27}+\frac{80}{81}-...\\ &\text{a}.\quad 5\qquad\qquad\qquad\:\: \text{d}.\quad 2\\ &\text{b}.\quad 4\qquad \text{c}.\quad 3\qquad \text{e}.\quad 1\\ \end{aligned}$
.Materi Barisan dan Deret
$\begin{aligned}46.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}=\:....\\ &\text{a}.\quad \displaystyle \frac{2}{3}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 3\\ &\text{b}.\quad 1\qquad\qquad \text{c}.\quad 2\qquad\qquad \text{e}.\quad 4\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Ingat bentuk}\\ & \underset{x\rightarrow \infty }{\textrm{Lim}}\: \sqrt{\displaystyle ax^{\displaystyle 2}+bx+c}-\sqrt{\displaystyle ax^{\displaystyle 2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga soal untuk di atas}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-(1+x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1}}\\ &=\displaystyle \frac{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1} \right)}{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1} \right)}\\ &=\displaystyle \frac{\left( \displaystyle \frac{-3-(-1)}{2\sqrt{1}} \right)}{\left( \displaystyle \frac{1-2}{2\sqrt{1}} \right)}=\displaystyle \frac{-2}{-1}=2\\\end{aligned}$
$\begin{aligned}47.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\:....\\ &\text{a}.\quad \displaystyle 1\:\:\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 9\\ &\text{b}.\quad 3\qquad\qquad \text{c}.\quad 4\qquad\qquad \text{e}.\quad 16\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x}.2+3^{\displaystyle x}.3}{\displaystyle \frac{2^{\displaystyle x}}{2}+\displaystyle \frac{3^{\displaystyle x}}{3}}\times \frac{\displaystyle \frac{1}{3^{\displaystyle x}}}{\displaystyle \frac{1}{3^{\displaystyle x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}\\ & \textrm{Perhatikan bahwa saat}\: x\longrightarrow \infty \:,\: \textrm{maka}\:\: \left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}\longrightarrow 0\\ &\textrm{Sehingga}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}= \displaystyle \frac{2.\left( 0 \right)+3}{\displaystyle \frac{1}{2}.\left( 0 \right)+\displaystyle \frac{1}{3}}=\displaystyle \frac{3}{\displaystyle \frac{1}{3}}=9 \end{aligned}$.
$\begin{aligned}48.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=\:....\\\\ &\textrm{dengan}\:\left\lfloor x \right\rfloor= \textrm{bilangan bulat terbesar yang }\\ &\textrm{kurang dari atau sama dengan}\:\:\: x\\\\ &\text{a}.\quad \displaystyle \frac{1}{2}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 1\\ &\text{b}.\quad 2\qquad\qquad \text{c}.\quad 0\qquad\qquad \text{e}.\quad \textrm{tidak ada}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Berdasarkan sifat fungsi tangga, untuk setiap bilangan}\\ &\textrm{riil}\:\:\: x\:\:\: \textrm{berlaku}:\\ & x-1<\left\lfloor x \right\rfloor\le x\\ &\text{Untuk}\quad x>0,\:\: \textrm{bagilah seluruh ruas dengan}\:\:\: x\\ &\displaystyle \frac{x-1}{x}<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le \displaystyle \frac{x}{x}\Leftrightarrow \left( 1-\displaystyle \frac{1}{x} \right)<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le 1\\ &\textrm{Selanjutnya kita hitung nilai limit batas kiri dan kanan}\\ &\text{ketika}:x\longrightarrow \infty \:(\textrm{ingat ini bukan limit kiri dan kanan})\\ &\circ \quad \textrm{Batas kiri}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( 1-\displaystyle \frac{1}{x} \right)=1-\displaystyle \frac{1}{\infty }=1-0=1\\ &\circ \quad \textrm{Batas kanan}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: 1=1\\ &\textrm{Berdasarkan teorema apit (Squeeze Theorem), karena}\\ &\textrm{batas kiri sama dengan batas kanan, maka nilai}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=1 \end{aligned} \end{aligned}$.
$\begin{array}{ll}\\ 41.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{18}{x\sin \displaystyle \frac{3}{x}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -54 \\ \textrm{b}.&\displaystyle -6\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{6}\\ \textrm{d}.&\displaystyle 6\\ \textrm{e}.&\displaystyle 54 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{18}{x\sin \displaystyle \frac{3}{x}}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18}{\displaystyle \frac{1}{u}\sin 3u}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18u}{\sin 3u}\\ &=\displaystyle \frac{18}{3}\\ &=6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 42.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x\sin \displaystyle \frac{2}{x}}{2}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4 \\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{2}\\ \textrm{d}.&\displaystyle 2\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x\sin \displaystyle \frac{2}{x}}{2}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{4}{u}.\sin 2u}{2}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{4\sin 2u}{2u}\\ &=\displaystyle \frac{4\times 2}{2}\\ &=4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 43.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=-\infty \end{aligned} \end{array}$
$\begin{array}{l}\\ 44.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: x=2\: \: \textrm{dan}\: \: x=3 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 45.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=-3 \end{aligned} \end{array}$.
DAFTAR PUSTAKA
$\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 37.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\infty \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 38.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle \frac{4}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle \frac{11}{4}\\ \textrm{e}.&\displaystyle 60 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle \frac{4}{x}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15}{u}.\tan 4u\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15\tan 4u}{u}\\ &=15\times 4\\ &=60 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 39.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=(ab)^{2} \end{aligned} \end{array}$
$\begin{array}{l}\\ 40.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=0 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 31.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 33.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{b}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{l}\\ 34.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=0 \end{aligned} \end{array}$
$\begin{array}{l}\\ 35.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=-2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 26.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=0 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 27.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\frac{7}{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 28.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 29.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=-\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 30.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{black}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=-\frac{5}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 21.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\infty \end{aligned} \end{array}$
$\begin{array}{l}\\ 22.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\sqrt{\infty }+\sqrt{\infty }\\ &=\infty \end{array}$
$\begin{array}{ll}\\ 23.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=-\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=2 \end{aligned} \end{array}$
$\begin{aligned}\textrm{ada cara lain yang lebih sede}&\textrm{rhana, yaitu:}\\ .\qquad\: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ \textrm{Jika}\quad &\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}&-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ \textrm{Sehingga}\quad&\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=2 \end{aligned}$
$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\sqrt{\infty }+\sqrt{\infty }=\infty \end{array}$
☝☝☝☝☝☝☝☝☝☝☝☝
$\begin{aligned}\textrm{Sebagai}&\: \: \textbf{CATATAN}\: \textrm{di sini}\\ \textrm{Sifat-sif}&\textrm{at bilangan tak hingga}\\ (1)\: \: &\infty +\infty =\infty \\ (2)\: \: &-\infty +(-\infty )=-\infty \\ (3)\: \: &\infty \times \infty =\infty\\ (4)\: \: &-\infty \times (-\infty )=\infty \\ (5)\: \: &k.\infty =\infty ,\quad k\: \: \textrm{positif}\\ (6)\: \: &k.(-\infty )=-\infty,\quad k\: \: \textrm{positif} \\ (7)\: \: &k.\infty =-\infty ,\quad k\: \: \textrm{negatif}\\ (8)\: \: &k.(-\infty )=\infty ,\quad k\: \: \textrm{negatif}\\ \textrm{yang ha}&\textrm{rus dihindari}\\ (1)\: \: &\infty -\infty ,\quad \: \: \textrm{bentuk tak tentu}\\ (2)\: \: &\displaystyle \frac{\infty }{\infty },\: -\displaystyle \frac{\infty }{\infty },\: \: \textrm{dan}\: \: \frac{0}{0} \end{aligned}$
$\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2026}{2x^{2}-2027x}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2026}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2027x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2026}{x^{2}}}{2-\displaystyle \frac{2027}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2026}{\infty ^{2}}}{2-\displaystyle \frac{2027}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2027}{\sqrt{9x^{2}-2028x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2027}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2028x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2027}{x}}{-\sqrt{9-\displaystyle \frac{2028}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2027}{\infty }}{-\sqrt{9-\displaystyle \frac{2028}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=25 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 19.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e} \end{array}$
$\begin{aligned}.\qquad \: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: &\: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=-\displaystyle \frac{5}{3} \end{aligned}$
$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=1 \end{aligned} \end{array}$
$\begin{aligned}11.\quad&\textrm{Perhatikan gambar grafik fungsi}\:\: y=f(x)\: \: \textrm{berikut}\end{aligned}$.
$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: a\in \left\{ 0,2,4 \right\}\:\: \textrm{dengan}\:\: \underset{x\rightarrow a }{\textrm{Lim}}\: f(x)\:\: \textrm{ada, maka}\\ &\textrm{untuk nilai semua}\:\: a\:\: \textrm{yang memenuhi adalah}\: ....\\ &\textrm{a}.\quad a=0\\ &\textrm{b}.\quad a=2\\ &\textrm{c}.\quad a=0\:\: \textrm{dan}\:\: a=2\\ &\textrm{d}.\quad a=0\:\: \textrm{dan}\:\: a=4\\ &\textrm{e}.\quad a=0\:,\: a=2\:,\: \textrm{dan}\:\: a=4\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas sesuai ilustrasi gambar, bahwa}\\ &\bullet \quad\underset{\displaystyle x\rightarrow 0^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow 0^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x)=6\\ &\textrm{dan}\\ &\bullet \quad \underset{\displaystyle x\rightarrow 2^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow 2^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x)=2\\ &\textrm{serta nilai}\\ &\bullet \quad \underset{\displaystyle x\rightarrow 4^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)\neq \underset{\displaystyle x\rightarrow 4^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x) \end{aligned}$.
$\begin{array}{l}\\ 12.&\textrm{Diketahui}\\ &f(x)=\begin{cases} 3x^{\displaystyle 2}+8x+1 & ,&\textrm{untuk}\quad x<2 \\ ax+3&,&\textrm{untuk}\quad x\ge 2 \end{cases}\\ &\textrm{Besar}\quad a\quad \textrm{yang memenuhi agar nilai}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: f(x)\:\: \textrm{ada}\quad\textrm{adalah}\: ....\\ &\begin{array}{llllllll} \textrm{a}.&-13\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 13\\ \textrm{b}.&-2\qquad\qquad \textrm{c}.\quad 2\:\quad\quad\qquad \textrm{e}.\quad 26\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Suatu fungsi}\quad f(x)\quad \textrm{memiliki nilai limit}\:\: \underset{x\rightarrow 2}{\textrm{Lim}}\: f(x)\\&\textrm{jika dan hanya jika nilai limit kiri sama dengan}\\ &\textrm{limit kanan pada}\quad x=2,\quad \textrm{yaitu}: \\ &\begin{aligned}&\underset{.\quad\displaystyle x\rightarrow 2^{\displaystyle -}}{\textrm{Lim}} f(x)\quad=\underset{.\quad\displaystyle x\rightarrow 2^{\displaystyle +}}{\textrm{Lim}}\: f(x)\\ &\Leftrightarrow 3(2)^{\displaystyle 2}+8(2)+1=a(2)+3\Leftrightarrow 29=2a+3\\ &\Leftrightarrow 2a=29-3\Leftrightarrow 2a=26\Leftrightarrow a=13\\ \end{aligned} \end{array}$.
$\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}$.
$\begin{array}{l}\\ 15.&\textrm{Dalam sebuah riset terkait peningkatan penggunaan}\\ &\textrm{data internet pada perangkat seluler, pertumbuhan}\\ &\textrm{penggunaan data (dalam gigabita) dimodelkanan}\\ &\textrm{oleh fungsi}\:\:f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{dengan}\:\:x\:\: \textrm{adalah}\\ &\textrm{waktu dalam bulan. Jika kecenderungan peningkatan}\\ &\textrm{waktu yang dihasilkan oleh fungsi}\quad f(x)\quad \textrm{terus}\\&\textrm{berlanjut, perkiraan penggunaan data mendekati}\\ &\textrm{tak hingga adalah}\: ....\: \textrm{gigabita}\\ &\begin{array}{llllllll} \textrm{a}.&300\:\:\qquad\qquad\qquad\qquad\quad\qquad \textrm{d}.\quad 100\\ \textrm{b}.&200\qquad\qquad \textrm{c}.\quad 150\:\quad\quad\qquad \textrm{e}.\quad 50\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Diketahui fungsi }\quad f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{untuk}:x\to \infty \\ & \textrm{maka}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: f(x)=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\times \displaystyle \frac{\displaystyle \frac{1}{x^{\displaystyle 2}}}{\displaystyle \frac{1}{x^{\displaystyle 2}}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100+\displaystyle \frac{200}{x}}{2+\displaystyle \frac{100}{x^{\displaystyle 2}}}\\&=\displaystyle \frac{100+\displaystyle \frac{200}{\infty }}{2+\displaystyle \frac{100}{\infty ^{\displaystyle 2}}}=\displaystyle \frac{100+0}{2+0}=50\\ \end{array}$.
