$\begin{array}{ll}\\ 21.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{b}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{l}\\ 22.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \color{red}\textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=\color{red}0 \end{aligned} \end{array}$
$\begin{array}{l}\\ 23.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\color{blue}\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=\color{red}-2 \end{aligned} \end{array}$.
$\begin{array}{l}\\ 24.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \color{red}\textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=\color{red}5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 25.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\color{red}\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 26.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \color{red}\textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=\color{red}(ab)^{2} \end{aligned} \end{array}$
$\begin{array}{l}\\ 27.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \color{red}\textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=\color{red}0 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 28.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\color{purple}\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$
$\begin{array}{l}\\ 29.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \color{red}\textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: \color{red}x=2\: \: \color{black}\textrm{dan}\: \: \color{red}x=3 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 30.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \color{red}\textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\color{blue}\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=\color{red}-3 \end{aligned} \end{array}$
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