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Question:
$\displaystyle\lim_{x\to0} \sqrt{\frac{x-\sin x}{x+\sin^{2}x}} $
is equal to
BITSAT - 2018
BITSAT
Updated On:
Jan 30, 2025
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$\infty$
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Solution and Explanation
$\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}}} $
$=\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 $
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