Question

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

• A. $\frac{1}{2}$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

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