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  • Evaluate  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x-\sin x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200%7D%

Evaluate  \mathrm{\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x-\sin x}}

Option: 1

6


Option: 2

3


Option: 3

2


Option: 4

1


Answers (1)

best_answer

Write

\mathrm{\frac{2 \sin x-\sin 2 x}{x-\sin x} =\frac{2 \sin x-2 \sin x \cos x}{x-\sin x}}
                                 \mathrm{=2\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^{2}}\right)\left(\frac{x^{3}}{x-\sin x}\right)}

Using standard limits
\mathrm{\lim _{x \rightarrow 0} \frac{\sin x}{x} =1}
\mathrm{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} =\frac{1}{2} }
\mathrm{\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}} =\frac{1}{6} }

it follows that
\mathrm{\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x-\sin x} =2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right) \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right) \lim _{x \rightarrow 0}\left(\frac{x^{3}}{x-\sin x}\right) }
                                          \mathrm{=2(1)\left(\frac{1}{2}\right)\left(\frac{6}{1}\right) }
                                          \mathrm{=6 }.

Posted by

shivangi.shekhar

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