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

$\begin{array}{l}\\ 1.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}=....}\\ \\ & \text{a}.{\displaystyle 8}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.2\\ & \text{d}.1\\ & \text{e}.{\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{8x^2}{x^2}+\frac{x}{x^2}-\frac{2020}{x^2}}}{{\displaystyle \frac{2x^2}{x^2}-\frac{2021x}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8+{\displaystyle \frac{1}{x}-\frac{2020}{x^2}}}{2-{\displaystyle \frac{2021}{x}}}}\\ & ={\displaystyle \frac{8+{\displaystyle \frac{1}{\mathrm{\infty }}-\frac{2020}{{\mathrm{\infty }}^2}}}{2-{\displaystyle \frac{2021}{\mathrm{\infty }}}}}\\ & ={\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2019}{\sqrt{9x^2-2020x}}=....}\\ \\ & \text{a}.{\displaystyle 3}\\ & \text{b}.{\displaystyle 1}\\ & \text{c}.{\displaystyle \frac{1}{3}}\\ & \text{d}.-{\displaystyle \frac{1}{3}}\\ & \text{e}.{\displaystyle -3}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{(-\sqrt{{\displaystyle \frac{1}{x^2}}})}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{x}{x}+\frac{2021}{x}}}{-\sqrt{{\displaystyle \frac{9x^2}{x^2}-\frac{2020x}{x^2}}}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1+{\displaystyle \frac{2021}{x}}}{-\sqrt{9-{\displaystyle \frac{2020}{x}}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{2021}{\mathrm{\infty }}}}{-\sqrt{9-{\displaystyle \frac{2020}{\mathrm{\infty }}}}}={\displaystyle \frac{1}{-\sqrt{9}}}}\\ & =-{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=....}\\ \\ & \text{a}.{\displaystyle 1}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.9\\ & \text{d}.16\\ & \text{e}.25\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times {\displaystyle \frac{{\displaystyle \frac{1}{5^x}}}{{\displaystyle \frac{1}{5^x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2{\left({\displaystyle \frac{2}{5}}\right)}^x+3{\left({\displaystyle \frac{3}{5}}\right)}^x+4{\left({\displaystyle \frac{4}{5}}\right)}^x+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{2}{5}}\right)}^x+\frac{1}{3}{\left({\displaystyle \frac{3}{5}}\right)}^x+\frac{1}{4}{\left({\displaystyle \frac{4}{5}}\right)}^x+\frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{0+0+0+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{0+0+0+{\displaystyle \frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{5.1}{{\displaystyle \frac{1}{5}.1}}}\\ & =25\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \mathbf{\text{(USM UGM Mat IPA)}}\\ & \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{5}{3}}\\ \\ \text{b}.{\displaystyle \frac{2}{3}}\\ \\ \text{c}.-{\displaystyle \frac{1}{3}}\\ \\ \text{d}.-{\displaystyle \frac{2}{3}}\\ \\ \text{e}.-{\displaystyle \frac{5}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\end{array}$

$\begin{array}{rl}.\underset{x\to \mathrm{\infty }}{\text{Lim}}& {\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-\sqrt[3]{{(x+1)}^3})}\\ & \text{ingat bentuk}a-b=(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})\\ & \text{dan untuk}\{\begin{array}{ll}a& =(x^3-2x^2)\\ & \\ b& ={(x+1)}^3\end{array}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-{(x+1)}^3}{\sqrt[3]{{(x^3-2x^2)}^2}+\sqrt[3]{(x^3-2x^2){(x+1)}^3}+\sqrt[3]{{\left({(x+1)}^3\right)}^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-(x^3+3x^2+3x+1)}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-5x^2+...}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & ={\displaystyle \frac{-5}{1+1+1}}\\ & =-{\displaystyle \frac{5}{3}}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai}\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 2}\\ \\ \text{d}.{\displaystyle \frac{5}{2}}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle ((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{k}-\frac{1}{k+1}))}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (1-\frac{1}{k+1})}\\ & ={\displaystyle (1-\frac{1}{\mathrm{\infty }+1})}\\ & ={\displaystyle 1-\frac{1}{\mathrm{\infty }}}\\ & =1-0\\ & =1\end{array}\end{array}$