Question

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

Answer 1

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