Latihan Soal 1 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll}1.& \text{Perhatikanlah pernyataan-pernyataan berikut}\\ & \text{a}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=8\\ & \text{b}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=4,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=4\\ & \text{c}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=4\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)\text{tidak ada}\\ & \text{d}.\text{Jika}\underset{x\to 0^{-}}{\text{Lim}}f(x)=3\text{atau}\underset{x\to 0^{+}}{\text{Lim}}f(x)=2,\\ & \text{maka}\underset{x\to 0}{\text{Lim}}f(x)=1\\ & \text{Pernyataan di atas yang tepat adalah}....\\ & \text{a}.(i)\text{dan}(ii)\\ & \text{b}.(i)\text{dan}(iii)\\ & \text{c}.(ii)\text{dan}(iii)\\ & \text{d}.(ii)\text{dan}(iv)\\ & \text{e}.(iii)\text{dan}(iv)\\ \\ & \mathbf{\text{Jawab}}:\text{c}\\ & \begin{array}{|c|}\hline \begin{array}{rl}\text{Ing}& \text{at}\mathbf{\text{Definisi Limit}}\text{berikut}:\\ \text{Mis}& \text{al}f\text{sebuah fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}c,L\in \mathbb{R}\\ 1.& \text{Limit fungsi trigonometri};\underset{x\to c}{\text{Lim}}f(x)=L\mathbf{\text{ada}}\\ & \text{untuk semua nilai}x\text{mendekati}c\\ & \text{jika dan hanya jika nilai}f(x)\text{mendekati}L\\ 2.& \underset{x\to c}{\text{Lim}}f(x)=L\iff \underset{x\to c^{-}}{\text{Lim}}f(x)=\underset{x\to c^{+}}{\text{Lim}}f(x)=L\\ & \text{Limit kiri}=\text{limit kanan}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{ll}2.& \text{Perhatikanlah gambar dan pernyatan-}\\ & \text{pernyataan berikut}\end{array}$.

$\begin{array}{ll}& \text{i}.\text{Nilai}\underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{Lim}}f(x)=4\\ & \text{ii}.\text{Nilai}\underset{x\to {\displaystyle \frac{3\pi }{4}}}{\text{Lim}}f(x)=\sqrt{2}\\ & \text{iii}.\text{Nilai}\underset{x\to \pi }{\text{Lim}}f(x)=1\\ & \text{iv}.\text{Nilai}\underset{x\to {\displaystyle \frac{5\pi }{4}}}{\text{Lim}}f(x)=-1\\ & \text{Pernyataan-pernyataan yang tepat}\\ & \text{ditunjukkan oleh}....\\ \\ & \text{a}.(i)\text{dan}(ii)\\ & \text{b}.(i)\text{dan}(iii)\\ & \text{c}.(ii)\text{dan}(iii)\\ & \text{d}.(ii)\text{dan}(iv)\\ & \text{e}.(iii)\text{dan}(iv)\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\end{array}$.

$\begin{array}{ll}3.& \text{Nilai}\underset{x\to 2}{\text{Lim}}f(x),\text{dengan kondisi}\\ & f(x)=\{\begin{array}{ll}{\displaystyle \frac{x^2-4}{x-2}}& \text{untuk}x\ne 2\\ & \\ 6x& \text{untuk}x=2\end{array}\\ & \text{adalah}....\\ & \text{a}.\text{tidak ada}\\ & \text{b}.0\\ & \text{c}.2\\ & \text{d}.4\\ & \text{e}.12\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui sebagaimana pada soal, maka}\\ & \text{Harga limit kiri}:\\ & \underset{x\to 2^{-}}{\text{Lim}}f(x)=\underset{x\to 2^{-}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{-}}{\text{Lim}}(x+2)=4}\\ & \text{Dan harga limit kanan}:\\ & \underset{x\to 2^{+}}{\text{Lim}}f(x)=\underset{x\to 2^{+}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{+}}{\text{Lim}}(x+2)=4}\\ & \text{Karena limit kiri}=\text{limit kanan},\text{yaitu}\\ & \underset{x\to 2^{-}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=\underset{x\to 2^{+}}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=4,\text{maka}}}\\ & \underset{x\to 2}{\text{Lim}}f(x)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^2-4}{x-2}=4}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 5.& \text{Nilai}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}=....}\\ \\ & \text{a}.{\displaystyle 3}\\ & \text{b}.{\displaystyle 1}\\ & \text{c}.{\displaystyle \frac{1}{3}}\\ & \text{d}.-{\displaystyle \frac{1}{3}}\\ & \text{e}.{\displaystyle -3}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{(-\sqrt{{\displaystyle \frac{1}{x^2}}})}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{x}{x}+\frac{2021}{x}}}{-\sqrt{{\displaystyle \frac{9x^2}{x^2}-\frac{2020x}{x^2}}}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1+{\displaystyle \frac{2021}{x}}}{-\sqrt{9-{\displaystyle \frac{2020}{x}}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{2021}{\mathrm{\infty }}}}{-\sqrt{9-{\displaystyle \frac{2020}{\mathrm{\infty }}}}}={\displaystyle \frac{1}{-\sqrt{9}}}}\\ & =-{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 6.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=....}\\ \\ & \text{a}.{\displaystyle 1}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.9\\ & \text{d}.16\\ & \text{e}.25\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times {\displaystyle \frac{{\displaystyle \frac{1}{5^x}}}{{\displaystyle \frac{1}{5^x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2{\left({\displaystyle \frac{2}{5}}\right)}^x+3{\left({\displaystyle \frac{3}{5}}\right)}^x+4{\left({\displaystyle \frac{4}{5}}\right)}^x+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{2}{5}}\right)}^x+\frac{1}{3}{\left({\displaystyle \frac{3}{5}}\right)}^x+\frac{1}{4}{\left({\displaystyle \frac{4}{5}}\right)}^x+\frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{0+0+0+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{0+0+0+{\displaystyle \frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{5.1}{{\displaystyle \frac{1}{5}.1}}}\\ & =25\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \mathbf{\text{(USM UGM Mat IPA)}}\\ & \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{5}{3}}\\ \\ \text{b}.{\displaystyle \frac{2}{3}}\\ \\ \text{c}.-{\displaystyle \frac{1}{3}}\\ \\ \text{d}.-{\displaystyle \frac{2}{3}}\\ \\ \text{e}.-{\displaystyle \frac{5}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\end{array}$

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

$\begin{array}{l}\\ 8.& \text{Nilai}\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 2}\\ \\ \text{d}.{\displaystyle \frac{5}{2}}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle ((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{k}-\frac{1}{k+1}))}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (1-\frac{1}{k+1})}\\ & ={\displaystyle (1-\frac{1}{\mathrm{\infty }+1})}\\ & ={\displaystyle 1-\frac{1}{\mathrm{\infty }}}\\ & =1-0\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{{\displaystyle \frac{1}{x}(\sqrt{8x-2020}+\sqrt{4x+2021})}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8x}{x^2}-\frac{2020}{x^2}}}+\sqrt{{\displaystyle \frac{4x}{x^2}+\frac{2021}{x^2}}})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8}{x}-{\displaystyle \frac{2020}{x}}}}+\sqrt{{\displaystyle \frac{4}{x}+{\displaystyle \frac{2021}{x}}}})}}\\ & ={\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{4}{0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\\ & =\sqrt{\mathrm{\infty }}+\sqrt{\mathrm{\infty }}\\ & =\mathrm{\infty }\end{array}$.

☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝

$\begin{array}{rl}\text{Sebagai}& \mathbf{\text{CATATAN}}\text{di sini}\\ \text{Sifat-sif}& \text{at bilangan tak hingga}\\ (1)& \mathrm{\infty }+\mathrm{\infty }=\mathrm{\infty }\\ (2)& -\mathrm{\infty }+(-\mathrm{\infty })=-\mathrm{\infty }\\ (3)& \mathrm{\infty }\times \mathrm{\infty }=\mathrm{\infty }\\ (4)& -\mathrm{\infty }+(-\mathrm{\infty })=\mathrm{\infty }\\ (5)& k.\mathrm{\infty }=\mathrm{\infty },k\text{positif}\\ (6)& k.(-\mathrm{\infty })=-\mathrm{\infty },k\text{positif}\\ (7)& k.\mathrm{\infty }=-\mathrm{\infty },k\text{negatif}\\ (8)& k.(-\mathrm{\infty })=\mathrm{\infty },k\text{negatif}\\ \text{yang ha}& \text{rus dihindari}\\ (1)& \mathrm{\infty }-\mathrm{\infty },\text{bentuk tak tentu}\\ (2)& {\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},-{\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},\text{dan}\frac{0}{0}}}\end{array}$.