Respuesta :

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[tex]\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}[/tex]

Notice that if [tex]f(x)=e^{\sin x}[/tex], then [tex]f(\pi)=e^{\sin\pi}=e^0=1[/tex]. Recall the definition of the derivative of a function [tex]f(x)[/tex] at a point [tex]x=c[/tex]:

[tex]f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]

So the value of this limit is exactly the value of the derivative of [tex]f(x)=e^{\sin x}[/tex] at [tex]x=\pi[/tex].

You have

[tex]f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1[/tex]

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