How do I prove this limit does not exist: $\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} $ i ve a doubt 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: For $x\to 0^+$,  we have that $e^{-1/x}\to 0$ (that is $e^{-\infty}$) and
$$\frac{e^{1/x} - 1}{e^{1/x} + 1}=\frac{1- e^{-1/x} }{1 + e^{-1/x} }\to 1$$
For $x\to 0^-$ we have that $e^{1/x}\to 0$ (again $e^{-\infty}$) and
$$\frac{e^{1/x} - 1}{e^{1/x} + 1}\to -1.$$
Left-limit and right-limit exist but they are different and the full limit does not exists. Therefore you can't use Hopital here.
A: L'hopital only applies if the limit exists, which this does not. Your reasoning about the left- and right-hand limits is correct. 
A: 
As you can see, and as already stated by the others, the limit from the right is $1$ whether the limit from the left is $-1$. I know some lecturers might tell you that you shouldn't plot the functions, but believe me it is so much better. When you want to evaluate a limit, the best thing, is to plot it and see indeed what happens and thinking of a way of going about showing that the limit exists (and in which case what it is) or that it doesn't (i.e. showing that you can find multiples limits or none at all).
So your idea of checking the limits on both sides is correct! 
