Answer
\[-1\]
Work Step by Step
To find:-
\[\lim_{x\rightarrow π}\frac{e^{\sin x}-1}{x-π}\]
Which is $\frac{0}{0}$ form
We will use L' Hopitals rule , which is , if $f(c)=g(c)=0$
\[\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\;\;\;...(1)\]
Using (1)
\[\lim_{x\rightarrow π}\frac{e^{\sin x}-1}{x-π}=\lim_{x\rightarrow π}\frac{(e^{\sin x}-1)'}{(x-π)'}\]
\[\lim_{x\rightarrow π}\frac{e^{\sin x}-1}{x-π}=\lim_{x\rightarrow π}\frac{\cos x\;e^{\sin x}}{1}\]
\[\Rightarrow \lim_{x\rightarrow π}\frac{e^{\sin x}-1}{x-π}=\cos π\;e^{\sin π}=-1\]
Hence , \[\lim_{x\rightarrow π}\frac{e^{\sin x}-1}{x-π}=-1\]
