$\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2(2 x)}$ is equal to
$\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2(2 x)}$ is equal to
- $0$
- $1$
- $4$
- $\frac{1}{4}$
Solution
$\begin{aligned} \lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cot ^2(2 x)} & =\lim _{x \rightarrow 0} \frac{x \tan ^2 2 x}{\sin ^2 x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{4\left(\frac{\tan 2 x}{2 x}\right)^2}{4\left(\frac{\sin x}{x}\right)^2\left(\frac{\tan 4 x}{4 x}\right)} \\ & =1\end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 2)
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