Q99 / 170 · Manipal (MET) 2018 · Mathematics

$\lim _{x \rightarrow 0}\left(\frac{(2+x) \sin (2+x)-2 \sin 2}{x}\right)$ is equal to

  • A
    $\sin 2$
  • B
    $\cos 2$
  • C
    1
  • Correct answer
    D
    $2 \cos 2+\sin 2$

Solution

$\begin{aligned} \lim _{x \rightarrow 0} & \frac{(2+x) \sin (2+x)-2 \sin 2}{x}\left(\frac{0}{0} \text { form }\right) \\ & =\lim _{x \rightarrow 0} \frac{\sin (2+x)+(2+x) \cos (2+x)}{1} \\ & =\sin 2+2 \cos 2\end{aligned}$

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

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