Question

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

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is A $1$

Solution:- (A) 1

$L={lim}_{x\to \infty }\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$

Let $x=\frac{1}{y}\Rightarrow \text{As}x\to \infty ,y\to 0$

$\therefore L={lim}_{y\to 0}\sqrt{\frac{\frac{1}{y}-\sin \left(\frac{1}{y}\right)}{\frac{1}{y}+{\cos }^2\left(\frac{1}{y}\right)}}={lim}_{y\to 0}\sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+y{\cos }^2\left(\frac{1}{y}\right)}}$

$\sin \left(\frac{1}{y}\right)$ and ${\cos }^2\left(\frac{1}{y}\right)$, both are bounded functions.

$\sin \left(\frac{1}{y}\right)\in [1,1]\text{for all}y\in R\text{and}{\cos }^2\left(\frac{1}{y}\right)\in [0,1]\text{for all}y\in R$.

$\Rightarrow {lim}_{y\to 0}y\sin \left(\frac{1}{y}\right)=0\text{and}{lim}_{y\to 0}y{\cos }^2\left(\frac{1}{y}\right)=0$

$\Rightarrow {lim}_{y\to 0}\sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+y{\cos }^2\left(\frac{1}{y}\right)}}={lim}_{y\to 0}\sqrt{\frac{1-0}{1+0}}=1$

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