$\begin{array}{ll}\\ 11.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \color{purple}\frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \color{red}\textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \color{purple}\displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=\color{red}2 \end{aligned} \end{array}$.
$\begin{aligned}&\textrm{ada cara lain yang lebih sede}\textrm{rhana, yaitu:}\\ .\qquad\: \, &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\qquad\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ &\textrm{Jika}\quad \\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga}\quad\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=\color{red}2 \end{aligned}$.
$\begin{array}{ll}\\ 13.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }=\color{red}\infty \end{array}$.
$\begin{array}{ll}\\ 14.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \color{purple}\displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=\color{red}0 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 15.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \color{red}\textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\color{red}\frac{7}{2} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 16.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 17.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \color{purple}\displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=\color{red}-\infty \end{aligned} \end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \color{red}\textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\color{blue}\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{purple}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=\color{red}-\frac{5}{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \color{purple}\displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\color{red}\displaystyle \frac{1}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \color{red}\textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\color{red}\infty \end{aligned} \end{array}$
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