Question

$\underset{x\to 0}{\lim }\sqrt{\frac{x-\sin x}{x+{\sin }^{2}x}}$ is equal to

• A. 1
• B. 0
• C. $\infty$
• D. None of these

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\sqrt{\frac{x-\sin x}{x+{\sin }^{2}x}}$
$=\underset{x\to 0}{\lim }\sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{{\sin }^{2}x}{x}}}$
$=\underset{x\to 0}{\lim }\sqrt{\frac{1-\frac{\sin x}{x}}{1+\left(\frac{\sin x}{x}\right)\sin x}}$
$=\sqrt{\frac{1-1}{1+1\times 0}}=0$