HOTS Zone : Limit Trigonometri
Table of Contents
Tipe:
No.
$\displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}=....$ALTERNATIF PENYELESAIAN
Misal $y=\left(\cot x\right)^\frac1{\ln x}$
$\begin{aligned} \ln y&=\ln\left(\left(\cot x\right)^\frac1{\ln x}\right)\\ &=\dfrac1{\ln x}\cdot\ln(\cot x)\\[3.7pt] &=\dfrac{\ln(\cot x)}{\ln x}\\[3.7pt] \displaystyle\lim_{x\to0^+}\ln y&=\displaystyle\lim_{x\to0^+}\dfrac{\ln(\cot x)}{\ln x}\\[3.7pt] \ln\displaystyle\lim_{x\to0^+}y&=\displaystyle\lim_{x\to0^+}\dfrac{-\dfrac1{\sin x\cos x}}{\dfrac1x}\\[3.7pt] &=\displaystyle\lim_{x\to0^+}\dfrac{-x}{\sin x\cos x}\\[3.7pt] &=\displaystyle\lim_{x\to0^+}\left(-\dfrac{x}{\sin x}\right)\cdot\displaystyle\lim_{x\to0^+}\dfrac1{\cos x}\\[3.7pt] &=(-1)\cdot\dfrac11\\[3.7pt] &=-1\\ \displaystyle\lim_{x\to0^+}y&=e^{-1}\\ \displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}&=\dfrac1e \end{aligned}$
$\begin{aligned} \ln y&=\ln\left(\left(\cot x\right)^\frac1{\ln x}\right)\\ &=\dfrac1{\ln x}\cdot\ln(\cot x)\\[3.7pt] &=\dfrac{\ln(\cot x)}{\ln x}\\[3.7pt] \displaystyle\lim_{x\to0^+}\ln y&=\displaystyle\lim_{x\to0^+}\dfrac{\ln(\cot x)}{\ln x}\\[3.7pt] \ln\displaystyle\lim_{x\to0^+}y&=\displaystyle\lim_{x\to0^+}\dfrac{-\dfrac1{\sin x\cos x}}{\dfrac1x}\\[3.7pt] &=\displaystyle\lim_{x\to0^+}\dfrac{-x}{\sin x\cos x}\\[3.7pt] &=\displaystyle\lim_{x\to0^+}\left(-\dfrac{x}{\sin x}\right)\cdot\displaystyle\lim_{x\to0^+}\dfrac1{\cos x}\\[3.7pt] &=(-1)\cdot\dfrac11\\[3.7pt] &=-1\\ \displaystyle\lim_{x\to0^+}y&=e^{-1}\\ \displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}&=\dfrac1e \end{aligned}$
Jadi, $\displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}=\dfrac1e$.
No.
Nilai dari $\displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}}$ yang paling mendekati adalah- 0
- 1
- −1
- −∞
- ∞
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}}&=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin^{4030}2x}{(4x)^{4030}}+\dfrac{\tan^{4030}4x}{(4x)^{4030}}}{\dfrac{\sin^{4030}4x}{(4x)^{4030}}-\dfrac{(2x)^{4030}}{(4x)^{4030}}}\\[3.7pt]
&=\displaystyle\lim_{x\to0}\dfrac{\dfrac1{2^{4030}}\cdot\dfrac{\sin^{4030}2x}{(2x)^{4030}}+\dfrac{\tan^{4030}4x}{(4x)^{4030}}}{\dfrac{\sin^{4030}4x}{(4x)^{4030}}-\dfrac1{2^{4030}}\cdot\dfrac{(2x)^{4030}}{(2x)^{4030}}}\\[3.7pt]
&=\dfrac{\dfrac1{2^{4030}}+1}{1-\dfrac1{2^{4030}}}\\[3.7pt]
&=\dfrac{2^{4030}+1}{2^{4030}-1}\\[3.7pt]
&=\dfrac{2^{4030}-1+2}{2^{4030}-1}\\[3.7pt]
&=1+\dfrac2{2^{4030}-1}\\
&=1+0{,}000...\\
&=1{,}000...\\
&\approx\boxed{\boxed{1}}
\end{aligned}
Jadi, nilai dari \displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}} yang paling mendekati adalah 1.
JAWAB: B
JAWAB: B
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, $\displaystyle\lim_{to+\infty}\left(5^{x-1}\tan\left(\dfrac3{5^x}\right)\right)=\dfrac35$.
JAWAB: A
JAWAB: A
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