Limit as $x\to 0$ with zero division: $ \lim_{x\to 0} {x\over e^x-e^{-x}} $ 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
 A: $$\begin{align}
  \mathop {\lim }\limits_{x \to 0} \frac{x}{{{e^x} - {e^{ - x}}}} &=\mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{{e^x}}}\frac{x}{{{e^x} - {e^{ - x}}}} \\ 
   =&\mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{1}\frac{x}{{{e^{2x}} - 1}} \\ 
   =& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{2}\frac{{2x}}{{{e^{2x}} - 1}} \\ 
   =& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{2}\mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{{e^{2x}} - 1}} \\ 
   =& \frac{{{e^0}}}{2} \cdot 1 \\ 
   = &\frac{1}{2} \end{align} $$
Note that $$\lim_{h\to 0}\frac{e^h-1}{h}=1$$is used in the fourth step.
A: 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}
$
A: HINT
$$\dfrac{x}{e^x - e^{-x}} = \dfrac{x}{(e^x-1) - (e^{-x}-1)} = \dfrac1{\dfrac{e^x-1}{x} - \dfrac{e^{-x}-1}{x}}$$
Recall what $\displaystyle \lim_{x \to 0} \dfrac{e^{ax} - 1}{x}$ is.
A: Have you learned L'Hopital's Rule yet? If so, since your limit is an indeterminate form of type $\dfrac 0 0$, we have $$\lim_{x\rightarrow 0} \frac{x}{e^x - e^{-x}}=\lim_{x\rightarrow 0} \frac{1}{e^x+e^{-x}}=\frac{1}{2}$$
