Q66 / 150 · MHT CET 2024 (16 May Shift 1) · Mathematics

The value of $\lim _{x \rightarrow 0}\left((\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$, where $x\gt0$ is

  • A
    0
  • B
    -1
  • Correct answer
    C
    1
  • D
    2

Solution

$\begin{aligned} & \lim _{x \rightarrow 0^0}\left\{(\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right\} \\ & =\lim _{x \rightarrow 0}(\sin x)^{\frac{1}{x}}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\ & =0+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\ & =\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}\end{aligned}$ Let $l=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}$. Then, $\begin{aligned} & \log l=\log \lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \\ & \Rightarrow \log l=\lim _{x \rightarrow 0}(-\sin x \log x) \\ & \Rightarrow \log l=-\lim _{x \rightarrow 0} \frac{\log x}{\operatorname{cosec} x} \\ & \Rightarrow \log l=-\lim _{x \rightarrow 0} \frac{\frac{1}{x}}{-\operatorname{cosec} x \cot x}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x \cos x} \\ & \Rightarrow \log l=\lim _{x \rightarrow 0} \frac{\tan x}{x} \times \sin x=1 \times 0=0 \\ & \Rightarrow l=\mathrm{e}^0=1 \end{aligned}$

Hence, the correct answer is option: 'C'.

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