_ x 0 2 x-x 3 x- x = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=$

A
$0$
B
1
✓
$\frac{1}{2}$
D
$\frac{1}{3}$

✓

Answer

Correct option: C.

$\frac{1}{2}$

(C)
$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0}\left\{\frac{\frac{2 \tan 2 x}{2 x}-1}{3-\frac{\sin x}{x}}\right\}=\frac{1}{2}$
Alternate method:
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{2 \sec ^2 2 x-1}{3-\cos x}$
$=\frac{2-1}{3-1}=\frac{1}{2}$

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