Q55 / 150 · MHT CET 2022 (07 Aug Shift 1) · Mathematics

$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=$

  • A
    $0$
  • Correct answer
    B
    $1$
  • C
    e
  • D
    $\frac{1}{\mathrm{e}}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=e^{\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}-1\right) \operatorname{cosec} x} \\ & =e^{\lim _{x \rightarrow 0}\left\{\frac{\tan x-\sin x}{1+\sin x} \cdot \frac{1}{\sin x}\right\}} \\ & \left.=e^{\lim _{x \rightarrow 0}\left\{\frac{\frac{1}{\cos x}-1}{1+\sin x}\right.}\right\} \\ & =\mathrm{e}^0=1 \\ & \end{aligned}$

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

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