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Q.
The value of $\underset{x \rightarrow 0^{+}}{\text{Lim}} \frac{e^{(\sin x)^x-1}-e^{x \ln (\sin x)}}{x \ln (\sin x)}$ is equal to
Continuity and Differentiability
Solution:
Let $(\sin x)^x=t$
$\text { as } x \rightarrow 0, t \rightarrow 1$
$\therefore \underset{ t \rightarrow 1} {\text{Lim}} \frac{ e ^{ t -1}- e ^{\ln t }}{\ln t } $
$\therefore e ^{\ln t } \operatorname{Lim}_{ t \rightarrow 1} \frac{ e ^{ t -1-\ln t }-1}{\ln t }=\underset{ t \rightarrow 1} {\text{Lim}} \frac{ t -1-\ln t }{\ln t }=\underset{ t \rightarrow 1} {\text{Lim}}\frac{1-\frac{1}{ t }}{\frac{1}{ t }}=0 .$