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I'm having trouble solving this limit:

$$ \lim_{x\to 0} {x\over e^x-e^{-x}} $$

Thanks for any help. I've tried expanding it by $(x+1)/(x+1)$, but it didn't help

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up vote 5 down vote accepted

$$\begin{align} \mathop {\lim }\limits_{x \to 0} \frac{x}{{{e^x} - {e^{ - x}}}} &=\mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{{e^x}}}\frac{x}{{{e^x} - {e^{ - x}}}} \\ =&\mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{1}\frac{x}{{{e^{2x}} - 1}} \\ =& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{2}\frac{{2x}}{{{e^{2x}} - 1}} \\ =& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{2}\mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{{e^{2x}} - 1}} \\ =& \frac{{{e^0}}}{2} \cdot 1 \\ = &\frac{1}{2} \end{align} $$

Note that $$\lim_{h\to 0}\frac{e^h-1}{h}=1$$is used in the fourth step.

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$$\dfrac{x}{e^x - e^{-x}} = \dfrac{x}{(e^x-1) - (e^{-x}-1)} = \dfrac1{\dfrac{e^x-1}{x} - \dfrac{e^{-x}-1}{x}}$$

Recall what $\displaystyle \lim_{x \to 0} \dfrac{e^{ax} - 1}{x}$ is.

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Nice, thank you! – Grant Nov 6 '12 at 23:59
@Grant Are you the same person( who posted this question? – user17762 Nov 7 '12 at 0:00
No. He's my classmate who asked me for help with this problem, but I didn't know the solution so I told him to try math.stackexchange. – Grant Nov 8 '12 at 2:03

As an alternative, this derivation implements a Big O notation for the exponential function.

$$ \displaystyle e^x=1+x+\frac{x^2}{2}+O\left(x^3\right) $$

$ \displaystyle L=\lim_{x\to 0} \, \frac{x}{\exp (x)-\exp (-x)}\\ \displaystyle L=\lim_{x\to 0} \, \frac{x}{\left(1+x+\frac{x^2}{2}+O\left(x^3\displaystyle \right)\right)-\left(1-x+\frac{x^2}{2}+O\left(x^3\right)\right)}\\ \displaystyle L=\lim_{x\to 0} \, \frac{x}{2 x+O\left(x^3\right)}\\ \displaystyle L=\lim_{x\to 0} \, \frac{1}{2+O\left(x^2\right)}\\ \displaystyle L=\frac{1}{2} $

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Have you learned L'Hopital's Rule yet? If so, since your limit is an indeterminate form of type $\dfrac 0 0$, we have $$\lim_{x\rightarrow 0} \frac{x}{e^x - e^{-x}}=\lim_{x\rightarrow 0} \frac{1}{e^x+e^{-x}}=\frac{1}{2}$$

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