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Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

1 Answer

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\displaystyle\lim_(x\to\pi)(e^(\sin x)-1)/(x-\pi)

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


f'(c):=\displaystyle\lim_(x\to c)(f(x)-f(c))/(x-c)

So the value of this limit is exactly the value of the derivative of
f(x)=e^(\sin x) at
x=\pi.

You have


f'(x)=\cos x\,e^(\sin x)\implies f'(\pi)=\cos\pi\,e^(\sin\pi)=-1
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