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Question: \(\underset{\mathbf{x}\mathbf{\rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\tan}\mathbf{2}\ma...

lim⁡x→0tan⁡2x−x3x−sin⁡x\underset{\mathbf{x}\mathbf{\rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\tan}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{x}}{\mathbf{3x}\mathbf{-}\mathbf{\sin}\mathbf{x}}equals

A

2/3

B

1/3

C

½

D

0

Answer

½

Explanation

Solution

lim⁡x→0tan⁡2x−x3x−sin⁡x=lim⁡x→0{2tan⁡2x2x−13−sin⁡xx}=12.\lim_{x \rightarrow 0}\frac{\tan 2x - x}{3x - \sin x} = \lim_{x \rightarrow 0}\left\{ \frac{\frac{2\tan 2x}{2x} - 1}{3 - \frac{\sin x}{x}} \right\} = \frac{1}{2}.