How do I prove this limit does not exist: $\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} $ How do I prove that this limit does not exist? 
$$\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} $$
My attempt:
When you approach from from left towards zero , say i take -0.00000000000001 . i substitute in expression i get (-1) . But if i take 0.000000000001 and substitue i get (=1) (by applying L'Hop) . But if i dont do this and apply L'Hop straighaway i get 1 . 
 A: The calculation of the limit between the lines below is wrong. L'Hopital's rule does not apply. See the edit at the end of this answer for the reason why.

We have
$$ \lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \rightarrow 0}\frac{f(x)}{g(x)}$$
where both of the following limits do not exist:
$$\lim_{x\rightarrow 0} f(x)$$
$$\lim_{x\rightarrow 0} g(x)$$
Thus we may apply L'Hôpital's rule to get 
$$\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}$$
We find the derivatives
$$f'(x)=g'(x)= -\frac{e^{1/x}}{x^2}$$
Since $f'(x) = g'(x)$ we notice $$\frac{f'(x)}{g'(x)}=1$$
This implies
$$\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}= \lim_{x \rightarrow 0} 1 = \boxed{1}$$

EDIT: I think L'Hôpital's rule actually doesn't apply, because of a subtlety in the conditions for its application. Looking at the wikipedia article, one of the conditions is that the limits of $f$ and $g$ are $\pm \infty$, but in your case the limit of $f$ and the limit of $g$ are both nonexistent (not infinity because on one side they are $\pm 1$ and on the other they are $\infty$). 
A: Look at limits from the right and left. The reason for the difference is that $\lim_{x\to0+}\frac1x=\infty$, while $\lim_{x\to0-}\frac1x=-\infty$:
\begin{align*}&\lim_{x\to0+}\frac{e^{1/x}-1}{e^{1/x}+1}=\lim_{x\to0+}\frac{1-\frac1{e^{1/x}}}{1+\frac1{e^{1/x}}}=\frac{1-\frac1\infty}{1+\frac1\infty}=\frac{1+0}{1-0}=1\\
&\lim_{x\to0-}\frac{e^{1/x}-1}{e^{1/x}+1}=\frac{e^{-\infty}-1}{e^{-\infty}+1}=\frac{0-1}{0+1}=-1\end{align*}
So the limit doesn't exist.
A: $$\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} $$
Seperate into two limits
$$\lim_{x\rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} $$
$$\lim_{x\rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} $$
Substitute $1/x=y$
$$\lim_{y\rightarrow \infty} \frac{e^{y} - 1}{e^{y} + 1} $$
$$\lim_{y\rightarrow -\infty} \frac{e^{y} - 1}{e^{y} + 1} $$
Substitute $e^y=z$
$$\lim_{z\rightarrow \infty} \frac{z - 1}{z + 1} $$
$$\lim_{z\rightarrow 0} \frac{z - 1}{z + 1} =\frac{0-1}{0+1}=-1$$
Therefore $\lim_{x\rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} =-1$
$$\lim_{z\rightarrow \infty} \frac{z - 1}{z + 1} =\frac\infty\infty$$
Use L'Hospital
$$\lim_{z\rightarrow \infty} \frac11=1$$
Therefore $\lim_{x\rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1}=1$
We have now shown that the left and right limits are not equal, therefore the limit does not exist.
A: $\displaystyle \lim_{x\rightarrow 0+} \frac{e^{1/x} - 1}{e^{1/x} + 1}= \displaystyle \lim_{x\rightarrow 0+} \tanh\left(\dfrac{1}{2x}\right)=1$
$\displaystyle \lim_{x\rightarrow 0-} \frac{e^{1/x} - 1}{e^{1/x} + 1}= \displaystyle \lim_{x\rightarrow 0-} \tanh\left(\dfrac{1}{2x}\right)=-1$
A: $\lim_{x\rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1}= \displaystyle \lim_{y\rightarrow \infty} \frac{e^{y} - 1}{e^{y} + 1} =1$ Note this result is the same as the result obtained using L'Hospital's rule 
$\lim_{x\rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1}= \displaystyle \lim_{y\rightarrow -\infty} \frac{e^{y} - 1}{e^{y} + 1} =-1$
They are not equal. So by definition the limit doesn't exist.
Also note L'Hospital's rule doesn't apply for the second limit,, since $\lim_{x\to 0^-}e^{1/x}=0$, so it's not the form for L'Hosptial's rule.
A: Let $$f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1} = 1 - \frac{2}{e^{1/x} + 1}.$$  Thus in order to characterize the limiting behavior of $f$ as $x \to 0$, it suffices to consider the behavior of $g(x) = e^{1/x}$ as $x \to 0$.  However it is obvious that $$\lim_{x \to 0^+} g(x) = \infty$$ whereas $$\lim_{x \to 0^-} g(x) = 0,$$ thus the two-sided limit fails to exist; consequently, the limit of $f$ at $0$ also fails to exist.  In any sufficiently small neighborhood of $0$, the value of $f$ for negative $x$ tends to $-1$, but for positive $x$, the value of $f$ tends to $1$.
The reason why L'Hopital's rule does not apply is simple:  as $x \to 0^-$, $f(x)$ is not an indeterminate form.
