Limit of exponential function of the form $\infty/\infty$ How would you solve for $$\lim_{x\to0} \frac{(e^\frac{1}{x}-1)}{(e^\frac{1}{x}+1)}$$
I tried to use the squeeze theorem, but this function doesn't have a definite range. 
 A: Let $x=1/u$:
$$L=\lim_{|u|\to+\infty}\frac{e^u-1}{e^u+1}$$
We may now see that as $u\to-\infty$, then by direct substition, we get
$$L_-=\frac{0-1}{0+1}=-1$$
But as $u\to+\infty$, we get
$$L_+=\lim_{u\to+\infty}\frac{e^u-1}{e^u+1}=\lim_{u\to+\infty}\frac{e^u+1-2}{e^u+1}=\lim_{u\to+\infty}1-\frac2{e^u+1}=1-0=1$$
So the limit does not exist.
A: A more symmetric treatment, for the connoisseurs: Multiplying by $\frac12 e^{-1/2x}$ above and below the fraction bar gives us
$$ \frac{e^{1/x}-1}{e^{1/x}+1} =
\frac{\frac12(e^{1/2x}-e^{-1/2x})}{\frac12(e^{1/2x}+e^{-1/2x})} =
\frac{\sinh(1/2x)}{\cosh(1/2x)} = \tanh(1/2x) $$
and since $\tanh(u)$ has different limits for $u\to+\infty$ and $u\to-\infty$, the original limit does not exist.
A: Notice that $$\forall x\in\mathbb{R}\backslash\{0\},\frac{e^\frac{1}{x}-1}{e^\frac{1}{x}+1}=1-\frac{2}{e^\frac{1}{x}+1}$$
A: Just simplify they fraction. 
$\displaystyle\frac{e^{1/x}-1}{e^{1/x}+1}=1-\frac{2}{e^{1/x}+1}$
then the answer is 1 if $x\to 0^+$
and when $x\to 0^-$ the answer is -1.
Therefore, the limit does not exist at x=0
