Question

Let $f(x)=(\sin x{)}^{\frac{1}{\pi -2x}},x\ne \frac{\pi }{2}$. If $f(x)$ is continuous at $x=\frac{\pi }{2}$ then find the value of $f\left(\frac{\pi }{2}\right)$.

Answer 1

Answer: 1.00

If function is continuous at $x=\pi /2$.
$f\left(\frac{\pi }{2}\right)=\underset{x\to \frac{\pi }{2}}{\lim }f(x)$
$=\underset{x\to \frac{\pi }{2}}{\lim }(\sin x{)}^{\frac{1}{\pi -2x}},1^{\infty }$
$=\underset{x\to \frac{\pi }{2}}{\lim }{e}^{\frac{1}{\pi -2x}(\sin x-1);\left(\frac{0}{0}\right)}$
$=\underset{x\to \frac{\pi }{2}}{\lim }{e}^{\frac{\cos x-0}{0-2};\text{ (by using L’ Hospital rule) }}$
$=e^{\frac{0}{-2}}=1$