Q82 / 200 · Manipal (MET) 2014 · Mathematics

$\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}$ is equal to

  • A
    0
  • B
    e
  • Correct answer
    C
    1
  • D
    Does not exist

Solution

$\begin{aligned} \lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x} & \\ & =\lim _{x \rightarrow 0}\left[\frac{\left(e^{\sin x}-1\right)}{\sin x} \times \frac{\sin x}{x}\right] \\ & =\lim _{\sin x \rightarrow 0} \frac{e^{\sin x}-1}{\sin x} \times \lim _{x \rightarrow 0} \frac{\sin x}{x} \\ & =1 \times 1=1\end{aligned}$

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

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