Q100 / 160 · TS EAMCET 2008 · Mathematics

$\lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}$ is equal to

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

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{\left(x+x^2\right) x} \\ & =\lim _{x \rightarrow 0} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots\right)}{x(1+x)} \times \lim _{x \rightarrow 0} \frac{\sin x}{x} \\ & =-1 \times 1=-1\end{aligned}$

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

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