Question

$\underset{x\to \infty }{lim}\frac{x-\sin x}{x+{\cos }^{2}x}$ equals

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

Answer 1

The correct option is B $1$

${lim}_{x\to \infty }\frac{x-\sin x}{x+{\cos }^{2}x}={lim}_{x\to \infty }\frac{x}{x}\left(\frac{1-\sin x/x}{1+{\cos }^{2}x/x}\right)$

as $x\to \infty \frac{1}{x}\to 0$

And we know that for any x

$-1\le \sin x\le 1,0\le {\cos }^{2}x\le 1$

$=\frac{1-\frac{\sin \left(\infty \right)}{\infty }}{1+\frac{{\cos }^2\left(\infty \right)}{\infty }}=\frac{1-0}{1+0}$

$=1$