Question:
$\displaystyle\lim_{x\to0} \sqrt{\frac{x-\sin x}{x+\sin^{2}x}} $ is equal to
$\displaystyle\lim_{x\to0} \sqrt{\frac{x-\sin x}{x+\sin^{2}x}} $ is equal to
Updated On: Dec 19, 2025
- 1
- 0
- $\infty$
- None of these
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The Correct Option is B
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 $
$=\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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