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Let \mathrm{f(x)=(\sin x)^{\frac{1}{\pi-2 x}}, x \neq \frac{\pi}{2}}    . If \mathrm{f(x)}  is continuous at  \mathrm{x=\frac{\pi}{2}}n , then  \mathrm{f\left(\frac{\pi}{2}\right)} is

Option: 1

e


Option: 2

1


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

Here, \mathrm{\lim _{h \rightarrow 0}\left\{\sin \left(\frac{\pi}{2}+h\right)\right\}^{\pi-2(\pi / 2+h)}=\lim _{h \rightarrow 0}\left\{\sin \left(\frac{\pi}{2}-h\right)\right\}^{\pi-2(\pi / 2-h)}=f\left(\frac{\pi}{2}\right)}

\mathrm{Now, \: \lim _{h \rightarrow 0}(\cos h) \frac{1}{(-2 h)}=e^{\lim _{h \rightarrow 0}(1 /(-2 h)) \log \cos h}=e^{\lim _{h \rightarrow 0} \frac{(1 / \cos h)(-\sin h)}{-2}}=e^0=1}

\mathrm{\therefore \quad f\left(\frac{\pi}{2}\right)=1 }

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