MCQ
$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to
  • A
    $\frac{1}{2}$
  • 1
  • C
    $0$
  • D
    -1

Answer

Correct option: B.
1
(B)
$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^3}$
$=2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{(1-\cos x)}{x^2}$
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$=2 \times 1 \times \frac{1^2}{2}$
$=1$

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