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  4. displaystyle lim x arrow (π/2) ( sin x-( sin x) sin x/1- sin x+ ln sin x) equals

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Q. $\displaystyle\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\ln \sin x}$ equals

Limits and Derivatives

Solution:

Let $\sin x=h$, then as $x \rightarrow \pi / 2, h \rightarrow 1$
$\therefore $ given limit
$=\displaystyle\lim _{h \rightarrow 1} \frac{h-h^{h}}{1-h+\ln h}$
$=\displaystyle\lim _{h \rightarrow 1} \frac{1-h^{h}-h^{h} \ln h}{-1+1 / h}$
(Using L’ Hospital Rule)
$=\displaystyle\lim _{h \rightarrow 1} \frac{-h^{h}-2 h^{h} \ln h-h^{h-1}-h^{h}(\ln h)^{2}}{-1 / h^{2}}$
$=\frac{-1-1}{-1}=2$