A Silly confusion about limits This is just a clarification about a question. My initial working of this question might be wrong. Please help me check

Question : Show that $\lim_{x \to 0}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}$ does not exist.

It is definitely clear that the left hand limit equals -1. 
For the right hand, $\lim_{x \to 0^+}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}$ = $\lim_{h \to 0}{\frac{e^{\frac{1}{h}} - 1}{e^{\frac{1}{h}} + 1}}$.  
Directly, substituting $ x = 0$ we get the limit as $\frac{\infty}{\infty}$ which does not exit. 
The above was my own thoughts which might be wrong.
But the limit can also be transformed into 
$\lim_{h \to 0}\frac{1 - \frac{1}{e^{1/h}}}{1 + \frac{1}{e^{1/h}}} = 1$. 
So my confusion is howcome when the right hand limit is not transformed it does not exist. But when transformed it exists as 1.
 A: As $x\to0^{-}$, $\frac{1}{x}\to -\infty$:
$$\lim_{x \to 0^{-}}e^{\frac{1}{x}}=e^{-\infty}=\frac{1}{e^{\infty}}=0$$
Therefore, the left-hand limit is:
$$\lim_{x \to 0^{-}}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}=
\frac{0-1}{0+1}=-1$$

As $x\to0^{+}$, $\frac{1}{x}\to\infty$:

$$\lim_{x \to 0^{+}}e^{\frac{1}{x}}=e^{\infty}=\infty$$
Therefore, the right-hand limit is:
$$\lim_{x \to 0^{+}}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}=
\lim_{x \to 0^{+}}{\frac{1 - \frac{1}{e^{\frac{1}{x}}}}{1 + \frac{1}{e^{\frac{1}{x}}}}}=
\frac{1 - \frac{1}{\infty}}{1 + \frac{1}{\infty}}=\frac{1-0}{1+0}=\frac{1}{1}=1
$$

Since the two one-sided limits are not equal, the limit itself does not exist:
$$
\lim_{x \to 0^{-}}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}\ne
\lim_{x \to 0^{+}}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}\implies
\lim_{x \to 0}{\frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}}=DNE
$$
A: Short answer: The limit does not exist because when you approach it from the left, it equals $-1$, but when you approach it from the right, it equals $1$. By definition, a two-sided limit does not exist when the two one-sided limits are not equal.
