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

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

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

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

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

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

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