limx→0tanx-xx2tanx equals to

$\lim_{x \to 0} \frac{\tan x - x}{x^{2} \tan x}$ equals to

A
1/3
B
2/3
C
1/2
D
1

Solution:

$\lim_{x \to 0} \frac{\tan x - x}{x^{2} \tan x} = \lim_{x \to 0} \frac{\sin x - x \cos x}{x^{2} \sin x} (\frac{0}{0} form)$

$= \lim_{x \to 0} \frac{\cos x - \cos x + x \sin x}{2 x \sin x + x^{2} \cos x}$

$(using L'Hôpital Rule)$

$= \lim_{x \to 0} \frac{\sin x}{2 \sin x + x \cos x} = \lim_{x \to 0} \frac{(\sin x) / x}{(2 (\sin x) / x) + \cos x} = \frac{1}{3}$

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