_ x 0 x-x x^3 = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^3}=$

A
$\frac{1}{3}$
B
$-\frac{1}{3}$
C
$\frac{1}{6}$
✓
$-\frac{1}{6}$

✓

Answer

Correct option: D.

$-\frac{1}{6}$

(D)
$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^3}$
$=\lim _{x \rightarrow 0} \frac{-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots}{x^3}$
$\ldots\left[\because \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots.\right]$
$=\lim _{x \rightarrow 0}\left(-\frac{1}{3!}+\frac{x^2}{5!}-\ldots\right)=\frac{-1}{3!}=\frac{-1}{6}$

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