Question

$\underset{x\to 2}{lim}\frac{\sqrt{4-x^2}\sqrt{\cos \frac{\pi x}{4}}}{\sin (x-2)}$ is

• A. $\sqrt{\frac{\pi }{4}}$
• B. $\sqrt{\frac{\pi }{2}}$
• C. $\sqrt{\frac{\pi }{8}}$
• D. $\sqrt{\frac{\pi }{16}}$

Answer 1

The correct option is B $\sqrt{\frac{\pi }{2}}$

${lim}_{x\to 2}\frac{\sqrt{4-x^2}\sqrt{\cos \frac{\pi x}{4}}}{\sin (x-2)}=l$ (say)

$\Rightarrow l={lim}_{x\to 2}\frac{\frac{\sqrt{4-x^2}\sqrt{\cos \frac{\pi x}{4}}}{(x-2)}}{\frac{\sin (x-2)}{(x-2)}}$ (Dividing numerator and denominator by $(x-2)$)

$\Rightarrow l=\frac{{lim}_{x\to 2}\frac{\sqrt{4-x^2}}{(x-2)}.\sqrt{\cos \frac{\pi }{4}x}}{{lim}_{x\to 2}\frac{\sin (x-2)}{(x-2)}}[\because {lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{{lim}_{x\to a}f\left(x\right)}{{lim}_{x\to a}g\left(x\right)}]$

$\Rightarrow l=\frac{{lim}_{x\to 2}\sqrt{\frac{(x+2)(2-x)}{{(2-x)}^2}}.\sqrt{\cos \frac{\pi }{4}x}}{1}[\because {lim}_{x\to a}\frac{\sin x}{x}=1]$

$\Rightarrow l={lim}_{x\to 2}\sqrt{\frac{x+2}{2-x}}.\sqrt{\cos \frac{\pi }{4}x}$

$\Rightarrow l={lim}_{x\to 2}\sqrt{\frac{(x+2)\left(\cos \frac{\pi }{4}x\right)}{2-x}}$

$\Rightarrow l=\sqrt{{lim}_{x\to 2}\frac{\left(x\right)\left(\cos \frac{\pi }{4}x\right)+2\cos \frac{\pi }{4}x}{2-x}}[\because {lim}_{x\to a}f\left(g\left(x\right)\right)=f\left({lim}_{x\to a}g\left(x\right)\right)]$

$\Rightarrow l=\sqrt{2}\sqrt{{lim}_{x\to 2}\frac{\cos \frac{\pi }{4}x}{2-x}}$

$\Rightarrow l=\sqrt{2}\sqrt{{lim}_{x\to 2}\frac{(-\sin \frac{\pi }{4}x)\left(\frac{\pi }{4}\right)}{-1}}$ ( by L-Hospital rule)

$\Rightarrow l=\sqrt{2}\sqrt{\frac{\frac{\pi }{2}\left(\sin \frac{\pi }{4}.2\right)}{1}}$

$\Rightarrow l=\sqrt{\frac{\pi }{4}.\left(2\right)\left(1\right)}$

$\Rightarrow l=\sqrt{\frac{\pi }{2}}$ is the correct answer