Dear student There is a printing mistake in the question Correct question is:limx→0tanx-sinxx(1+cos2x)=limx→0tanx-sinx2xcos2x=12limx→0tanxx-sinxxcos2x=12limx→0tanxx-sinxxlimx→0cos2x=12×01=0 Regards

Solution...

Dear student

There is a printing mistake in the question . Correct question is : \lim_{x \to 0} \frac{tanx - sinx}{x (1 + \cos 2 x)} = \lim_{x \to 0} \frac{tanx - sinx}{2 {xcos}^{2} x} = \frac{1}{2} \lim_{x \to 0} \frac{\frac{tanx}{x} - \frac{sinx}{x}}{\cos^{2} x} = \frac{1}{2} \frac{\lim_{x \to 0} (\frac{tanx}{x} - \frac{sinx}{x})}{\lim_{x \to 0} \cos^{2} x} = \frac{1}{2} \times \frac{0}{1} = 0

Regards

-7

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