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# calculus - What is $\lim_{x \to \pi}\frac{e^{\sin x} - 1}{x - \pi}$?

Please don't give me the answer - I only want a hint. $$\lim_{x \to \pi}\frac{e^{\sin x} - 1}{x - \pi}$$ This is a "Problems Plus" question from Stewart's Early Transcendentals (specifically, chapter 3 question 15).

I have no idea how to solve this limit - I didn't even think it existed at first because the limit of the numerator is $-1$ and the limit of the denominator is $0$, but Wolfram Alpha goes against this logic in saying that the limit is $-1$.

I've tried transforming this limit into the definition of a derivative for an easier time but have failed.

Any help is appreciated.

EDIT: This was in a section before L'Hoptial's Rule. However, I just realized the answer to the question - my apologies, I kind of forgot $e^0 = 1$ and not $0$.

1. 2019-11-14

$$\lim_{x\to\pi}\dfrac{e^{\sin(x)}-e^{\sin \pi}}{x-\pi}$$ Do you recognize the derivative's limit definition here? (The answer is $\cos \pi\times e^{\sin \pi}=-1$.)

2. 2019-11-14

Hint $$f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$

3. 2019-11-14

Use substitution $x-\pi=t$, then $t \to 0$. Then the limit becomes $$\lim_{t \to 0}\frac{e^{-\sin t}-1}{t}.$$ Now you may use series expansion and the fact $\frac{\sin t}{t} \to 1$ as $t \to 0$ to get to the actual limit.