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

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

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

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

$\begin{array}{ll} 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x),\: \: \textrm{dengan kondisi}\\ &\qquad\qquad f(x)=\begin{cases} \displaystyle \frac{x^{2}-4}{x-2} &\textrm{untuk}\: \: x\neq 2 \\ &\\ 6x & \textrm{untuk}\: \: x=2 \end{cases}\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{tidak ada}\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 4\\ &\textrm{e}.\quad 12\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sebagaimana pada soal, maka}\\ &\textrm{Harga limit kiri}:\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Dan harga limit kanan}:\\ &\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: (x+2)=4\\ &\textrm{Karena limit kiri}=\textrm{limit kanan},\: \: \textrm{yaitu}\\ &\underset{x\rightarrow 2^{-} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\underset{x\rightarrow 2^{+} }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=4,\: \: \textrm{maka}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: f(x)=\underset{x\rightarrow 2 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-4}{x-2}=\color{red}4 \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\color{red}\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\times \color{purple}\displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2021}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2020x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2021}{x}}{-\sqrt{9-\displaystyle \frac{2020}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2021}{\infty }}{-\sqrt{9-\displaystyle \frac{2020}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=\color{red}-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\color{red}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \color{purple}\displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=\color{red}25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \color{red}\textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{e} \end{array}$

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

$\begin{array}{ll}\\ 8.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \color{red}\textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=\color{red}1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \color{purple}\frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\color{red}\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \color{red}\textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\color{blue}\sqrt{\infty }+\sqrt{\infty }\\ &=\color{red}\infty \end{array}$.

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$\color{purple}\begin{aligned}\textrm{Sebagai}&\: \: \color{black}\textbf{CATATAN}\: \textrm{di sini}\\ \textrm{Sifat-sif}&\textrm{at bilangan tak hingga}\\ (1)\: \: &\infty +\infty =\infty \\ (2)\: \: &-\infty +(-\infty )=-\infty \\ (3)\: \: &\infty \times \infty =\infty\\ (4)\: \: &-\infty +(-\infty )=\infty \\ (5)\: \: &k.\infty =\infty ,\quad k\: \: \color{blue}\textrm{positif}\\ (6)\: \: &k.(-\infty )=-\infty,\quad k\: \: \color{blue}\textrm{positif} \\ (7)\: \: &k.\infty =-\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ (8)\: \: &k.(-\infty )=\infty ,\quad k\: \: \color{red}\textrm{negatif}\\ \textrm{yang ha}&\textrm{rus dihindari}\\ (1)\: \: &\color{red}\infty -\infty ,\quad \: \: \color{black}\textrm{bentuk tak tentu}\\ (2)\: \: &\color{red}\displaystyle \frac{\infty }{\infty },\: -\displaystyle \frac{\infty }{\infty },\: \: \color{black}\textrm{dan}\: \: \color{red}\frac{0}{0} \end{aligned}$.