Question

If $f\left(x\right)=\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$, then $\underset{x\to \infty }{lim}f\left(x\right)$ is

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

Answer 1

Answer: (B)

$1$

$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to \infty }{lim}\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$
$=\underset{x\to \infty }{lim}\sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{{\cos }^{2}x}{x}}}$
$=\sqrt{\frac{1-0}{1+0}}$
$[\because \frac{\sin x}{x}\to 0,\frac{{\cos }^{2}x}{x}\to 0asx\to \infty ]$
$=1$