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

(A) 0

(B) 1

(C) $1 / 2$

(D) $-1 / 2$

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Best answer

SOLUTION —

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

$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{\tan x(1-\cos x)}{x^3} \\=\lim _{x \rightarrow 0} \frac{\tan x}{x} \cdot \frac{2 \sin ^2(x / 2)}{4 \cdot(x / 2)^2}=\frac{1}{2}\end{array}$

So, The correct option will be (C).

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