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Mathematics Question on limits of trigonometric functions

lim⁡x→0x−sin⁡xx+sin⁡2x\displaystyle\lim_{x\to0} \sqrt{\frac{x-\sin x}{x+\sin^{2}x}} is equal to

A

1

B

0

C

∞\infty

D

None of these

Answer

0

Explanation

Solution

lim⁡x→0x−sin⁡xx+sin⁡2x=lim⁡x→01−sin⁡xx1+sin⁡2xx\displaystyle\lim_{x\to0} \sqrt{\frac{x-\sin x}{x+\sin^{2}x}} = \displaystyle\lim_{x\to0} \sqrt{\frac{1- \frac{\sin x}{x}}{1+ \frac{\sin^{2}x}{x}}}
=lim⁡x→01−sin⁡xx1+(sin⁡xx)sin⁡x=1−11+1×0=0=\displaystyle\lim_{x\to0} \sqrt{\frac{1- \frac{\sin x}{x}}{1+ \left(\frac{\sin x}{x}\right)\sin x}} = \sqrt{\frac{1-1}{1+1\times0}} = 0