Let $f(x)=(\sin x)^{\frac{1}{π-2x}},\,x≠\frac{π}{2}$. If f (x) is continuous at $x=\frac{π}{2}$, then $f(\frac{π}{2})$ is

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Here, $\underset{h→0}{\lim}\begin{Bmatrix}\sin\begin{pmatrix}\frac{π}{2}+h\end{pmatrix}\end{Bmatrix}^{\frac{1}{π-2(π/2+h)}}=\underset{h→0}{\lim}\begin{Bmatrix}\sin\begin{pmatrix}\frac{π}{2}-h\end{pmatrix}\end{Bmatrix}^{\frac{1}{π-2(π/2-h)}}=f\begin{pmatrix}\frac{π}{2}\end{pmatrix}$

Now, $\underset{h→0}{\lim}(\cos h)^{\frac{1}{(-2h)}}=e^{\underset{h→0}{\lim}(1/(-2h))\log\cos h}=e^{\underset{h→0}{\lim}\frac{(1/\cos h)(-\sin h)}{2}}=e^0=1$

$∴f\begin{pmatrix}\frac{π}{2}\end{pmatrix}=1$