Question

The value of $\underset{x\to \pi /2}{lim}\frac{\sin (x\cos x)}{\cos \left(x\sin x\right)}$ is equal to

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

Answer 1

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

$L=\underset{x\to \pi /2}{lim}\frac{\sin \left(x\cos x\right)}{sin\left(\frac{\pi }{2}-x\sin x\right)}$
$=\underset{x\to \pi /2}{lim}\frac{\sin \left(x\cos x\right)}{\left(x\cos x\right)}\frac{x\cos x}{sin(\frac{\pi }{2}-x\sin x)}\frac{\left(\frac{\pi }{2}-x\sin x\right)}{\left(\frac{\pi }{2}-x\sin x\right)}$

$=1\times 1\underset{x\to \pi /2}{lim}\frac{x\cos x}{\left(\frac{\pi }{2}-x\sin x\right)}$
$Putx=\pi /2+h$

$L=\underset{h\to 0}{lim}\frac{\left(\frac{\pi }{2}+h\right)cos\left(\frac{\pi }{2}+h\right)}{\frac{\pi }{2}-\left(\frac{\pi }{2}+h\right)\sin \left(\frac{\pi }{2}+h\right)}$

$=\underset{h\to 0}{lim}\frac{-\left(\frac{\pi }{2}+h\right)\sin h}{\frac{\pi }{2}(1-\cos h)-h\cos h}$

$=\underset{h\to 0}{lim}\frac{-\left(\frac{\pi }{2}+h\right)\left(\frac{\sin h}{h}\right)}{\frac{\pi }{2}\frac{(1-\cos h)}{h}-\cos h}\text{(Divide Numerator and Denominator by h)}$

$=\frac{-\left(\frac{\pi }{2}+0\right)1}{0-1}=\frac{\pi }{2}$