Question

If $\underset{x\to \frac{\pi }{2}}{lim}\frac{\sin x}{x}=l$ and $\underset{x\to \infty }{lim}\frac{\cos x}{x}=m$, then which one of the following is correct?

• A. $l=1,m=1$
• B. $l=\frac{2}{\pi },m=\infty$
• C. $l=\frac{2}{\pi },m=0$
• D. $l=1,m=\infty$

Answer 1

The correct option is B $l=\frac{2}{\pi },m=\infty$

$l={lim}_{x\to \frac{\pi }{2}}\frac{sinx}{x}={lim}_{h\to 0}\frac{sin(\frac{\pi }{2}+h)}{\frac{\pi }{2}+h}$

$l={lim}_{h\to 0}\frac{sin\left(\frac{\pi }{2}\right)cosh+cos\frac{\pi }{2}sinh}{\frac{\pi }{2}+h}$

$l={lim}_{h\to 0}\frac{1.cosh+0.sinh}{\frac{\pi }{2}+h}$

$l=\frac{1+0}{\frac{\pi }{2}+0}=\frac{2}{\pi }$

$m={lim}_{x\to \infty }\frac{cosx}{x}$

$cosx$ will always remain between -1 and 1 for any value $x$

So, m = $\frac{[-1,1]}{\infty }=0$ ..... Option (B)