Q106 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Mathematics

$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to

  • Correct answer
    A
    1
  • B
    0
  • C
    -1
  • D
    does not exist

Solution

Let $L=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}\left(\right.$ form $\left.\frac{0}{0}\right)$ (using L-Hospital rule) $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{3 x^2}\left(\text { form } \frac{0}{0}\right) \\ & \text { (using L-Hospital rule) } \\ & =\lim _{x \rightarrow 0} \frac{-2 \sin x+4 \sin 2 x}{6 x}\left(\text { form } \frac{0}{0}\right) \\ & \quad \begin{array}{l}\text { (using L-Hospital rule) } \\ =\lim _{x \rightarrow 0} \frac{-2 \cos x+8 \cos 2 x}{6}=\frac{-2+8}{6}=1\end{array}\end{aligned}$

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

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