# Evaluating a limit involving trigonometry

I really thank you for your answers to my first question--I could easily solve first problem and a few more ones without another question.

But a while later I got another one while studying, then I attempted to solve it on my own. Despite my efforts to solve I have spent scores of minutes with neither finding one correct solution nor making any process.

Here is the problem:

$$\lim_{x \rightarrow 0} \frac{e^{1-\sin x}-e^{1-\tan x}}{\tan x-\sin x}$$

The happiness would be mine, if you could let me know how to solve or even a few hints.

Cheers.

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In this case you can use the mean value theorem to see that there is a $y$ in the interval $[1-\tan x,1-\sin x]$ such that

$$\frac{e^{1-\sin x}-e^{1-\tan x}}{\tan x -\sin x}=e^y$$ so the expression limits to $e^1=e$

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The expression limits to $e$, not $1$. –  nbubis Jul 6 '14 at 0:05
oh sorry, thanks. –  Rene Schipperus Jul 6 '14 at 0:06

Hint:

Use the expansion $\exp(1-x) \approx e - ex + O(x^2)$

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HINT:

$$\frac{e^{1-\sin x}-e^{1-\tan x}}{\tan x-\sin x}$$

$$=e\cdot\frac{e^{(\tan x-\sin x)}-1}{\tan x-\sin x}\cdot\frac1{e^{\tan x}}$$

Use $\displaystyle\lim_{h\to0}\frac{e^{(h)}-1}h=1$

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