Question

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

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

Answer 1

Answer: (D)

$\underset{x\to \infty }{\lim }\frac{\sin x}{x}=0,\underset{x\to \infty }{\lim }\frac{\cos x}{x}=0$
Let $A=\underset{x\to \infty }{\lim }\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$
$\Rightarrow$ $A=\underset{x\to \infty }{\lim }\sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{{\cos }^{2}x}{x}}}$
$\Rightarrow$ $A=\sqrt{\frac{1-0}{1+0}}=1$