HOTS Zone : Limit Tak Hingga
Table of Contents
Tipe:
No.
- −4
- −2
- 1
- 2
- 3
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to\infty}\left(\left(\dfrac12\right)^{3x}+\left(\dfrac12\right)^x\right)^{\frac1{x^2}}&=\displaystyle\lim_{x\to\infty}\left(\dfrac1{2^{3x}}+\dfrac1{2^x}\right)^{\frac1{x^2}}\\[3.7pt]
&=\displaystyle\lim_{x\to\infty}\left(\dfrac{2^{2x}+1}{2^{3x}}\right)^{\frac1{x^2}}\\[3.7pt]
&=\displaystyle\lim_{x\to\infty}\left(\dfrac{2^{2x}}{2^{3x}}\right)^{\frac1{x^2}}\\[3.7pt]
&=\displaystyle\lim_{x\to\infty}\left(2^{-x}\right)^{\frac1{x^2}}\\[3.7pt]
&=\displaystyle\lim_{x\to\infty}2^{-\frac1x}\\[3.7pt]
&=2^{-0}\\[6pt]
&=\boxed{\boxed{1}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\left(\dfrac12\right)^{3x}+\left(\dfrac12\right)^x\right)^{\frac1{x^2}}=1 .
No.
\[\lim_{n\to\infty}\dfrac{\sqrt[n]8-1}{\sqrt[n]{16}-1}=....\]ALTERNATIF PENYELESAIAN
\(\displaystyle\lim_{n\to\infty}\dfrac{\sqrt[n]8-1}{\sqrt[n]{16}-1}=\displaystyle\lim_{n\to\infty}\dfrac{\left(\sqrt[n]2\right)^3-1}{\left(\sqrt[n]2\right)^4-1}\)
let \(\sqrt[n]2=p\to1\)
\(\begin{aligned} \lim_{n\to\infty}\dfrac{\left(\sqrt[n]2\right)^3-1}{\left(\sqrt[n]2\right)^4-1}&=\lim_{p\to1}\dfrac{p^3-1}{p^4-1}\\ &=\lim_{p\to1}\dfrac{(p-1)\left(p^2+p+1\right)}{(p-1)(p+1)\left(p^2+1\right)}\\[4pt] &=\lim_{p\to1}\dfrac{p^2+p+1}{(p+1)\left(p^2+1\right)}\\[4pt] &=\dfrac{1^2+1+1}{(1+1)\left(1^2+1\right)}\\ &=\boxed{\boxed{\dfrac34}} \end{aligned}\)
let \(\sqrt[n]2=p\to1\)
\(\begin{aligned} \lim_{n\to\infty}\dfrac{\left(\sqrt[n]2\right)^3-1}{\left(\sqrt[n]2\right)^4-1}&=\lim_{p\to1}\dfrac{p^3-1}{p^4-1}\\ &=\lim_{p\to1}\dfrac{(p-1)\left(p^2+p+1\right)}{(p-1)(p+1)\left(p^2+1\right)}\\[4pt] &=\lim_{p\to1}\dfrac{p^2+p+1}{(p+1)\left(p^2+1\right)}\\[4pt] &=\dfrac{1^2+1+1}{(1+1)\left(1^2+1\right)}\\ &=\boxed{\boxed{\dfrac34}} \end{aligned}\)
Jadi,
JAWAB:
JAWAB:
No.
$\displaystyle\lim_{x\to+\infty}\left(5^{x-1}\tan\left(\dfrac3{5^x}\right)\right)$ = ....- $\dfrac35$
- $\dfrac3{10}$
- $\dfrac3{15}$
- $\dfrac3{20}$
- $\dfrac3{25}$
ALTERNATIF PENYELESAIAN
Misal 5−x = y
Jika x → +∞ maka y → 0
\(\begin{aligned} \displaystyle\lim_{x\to+\infty}\left(5^{x-1}\tan\left(\dfrac3{5^x}\right)\right)&=\displaystyle\lim_{x\to{+\infty}}\dfrac{\tan\left(3\cdot5^{-x}\right)}{5^{-x+1}}\\ &=\displaystyle\lim_{x\to{+\infty}}\dfrac{\tan\left(3\cdot5^{-x}\right)}{5\cdot5^{-x}}\\ &=\displaystyle\lim_{y\to0}\dfrac{\tan\left(3y\right)}{5y}\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac35}} \end{aligned}\)
Jika x → +∞ maka y → 0
\(\begin{aligned} \displaystyle\lim_{x\to+\infty}\left(5^{x-1}\tan\left(\dfrac3{5^x}\right)\right)&=\displaystyle\lim_{x\to{+\infty}}\dfrac{\tan\left(3\cdot5^{-x}\right)}{5^{-x+1}}\\ &=\displaystyle\lim_{x\to{+\infty}}\dfrac{\tan\left(3\cdot5^{-x}\right)}{5\cdot5^{-x}}\\ &=\displaystyle\lim_{y\to0}\dfrac{\tan\left(3y\right)}{5y}\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac35}} \end{aligned}\)
Jadi, $\displaystyle\lim_{x\to+\infty}\left(5^{x-1}\tan\left(\dfrac3{5^x}\right)\right)=\dfrac35$.
JAWAB: A
JAWAB: A
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