Why isn't L'Hospital's Rule applicable to $\lim_{x \to 0} \frac{e^{1/x}}{e^{1/x}+1}$? Why isn't L'Hospital's Rule applicable to $\displaystyle\lim_{x \to 0} \frac{e^{1/x}}{e^{1/x}+1}$?
I'm trying to use this form of LH Rule:

Let $f$ and $g$ be functions continuous and differentiable on $(a,b)$ and suppose that $$\lim_{x\to a^+} f(x) = \lim_{x\to a^+} g(x) = \infty.$$
  If $g'(x) \ne 0$, and if $\lim\limits_{x\to a^+} f'(x)/g'(x)$ exists (finite or infinite), then $$\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x\to a^+} \frac{f'(x)}{g'(x)}.$$

 A: L'Hopital's rule does not apply because the limit
$$
\lim_{x\to 0}e^{1/x}
$$
does not exist, since
$$
\lim_{x\to 0^-}e^{1/x}=0\quad\text{while}\quad \lim_{x\to 0^+}e^{1/x}=\infty
$$
So, consequently
$$
\lim_{x\to 0^-}\frac{e^{1/x}}{e^{1/x}+1}=0\quad\text{while}\quad \lim_{x\to 0^+}\frac{e^{1/x}}{e^{1/x}+1}=1
$$
A: You can apply l’Hôpital for the limit from the right, because it is in the form $\infty/\infty$.
The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $\infty/\infty$ on both sides.
More generally you need $\infty/\infty$ or $0/0$ on either side.
If you blindly apply l’Hôpital, you end up with
$$
\lim_{x\to0}\frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
\lim_{x\to0^-}\frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
\lim_{x\to0^+}\frac{e^{1/x}}{e^{1/x}+1}=\lim_{x\to0^+}\frac{1}{1+e^{-1/x}}=
\frac{1}{1+0}=1
$$
A: here, $$when~~~x\to 0^+,e^{\frac{1}{x}}\to \infty$$
So,$$\lim_{x \to 0^+} \frac{e^\frac{1}{x}}{e^\frac{1}{x}+1}=\lim_{x \to 0^+}\frac{e^\frac{1}{x}}{e^\frac{1}{x}\left(1+\frac{1}{e^\frac{1}{x}} \right)}=\lim_{x \to 0^+}\frac{1}{\left(1+\frac{1}{e^\frac{1}{x}} \right)}=\frac{1}{1+0}=1$$
$$when~~~x\to 0^-,e^{\frac{1}{x}} \to 0$$
So,
$$\lim_{x \to 0^-} \frac{e^\frac{1}{x}}{e^\frac{1}{x}+1}=\frac{0}{0+1}=0$$
so,simply limit doesn't exist.so we can't find the limit when $x \to 0$ no matter what procedure you follow.
