# Why doesn't the limit of $\frac{e^{\frac1x}-1}{e^{\frac1x}+1}$ exist?

Consider $$\lim_{x \to 0} \frac{e^{\frac1x}-1}{e^{\frac1x}+1}$$

Applying L'hospital's rule for the left hand limit and right hand limit gives the same answer.

Why doesn't this limit exist?

• Is $f$ that quotient or is that quotient the result of the limit? Commented Apr 13, 2017 at 18:04
• HINT.-$\dfrac{e^{\frac1x}-1}{e^{\frac1x}+1}=\dfrac{e^{\frac1x}+1-2}{e^{\frac1x}+1}=1-\dfrac{2}{e^{\frac1x}+1}$ but $e^{\frac 1x}$ tends to $0$ when $x$ tends to $0$ by the left and to $\infty$ when by yhe right. It follows the the limit of the function "tends" to $\pm1$. Commented Nov 1, 2022 at 14:23

You can't use L'Hopital for the left hand limit, because it is not of the form $\frac{\infty}{\infty}$ or any other indeterminate form.
$$\lim_{x\to0^-} e^{1/x} = 0$$
So $$\lim_{x\to 0^-}f(x)=\frac{-1}{1}$$
You can use L'Hopital to show that $\lim_{x\to 0^+} f(x)=1$.
• Yes, L'Hopital can be applied to one-sided limits. You can think of it via the equality $$\lim_{x\to 0^{+}} f(x)=\lim_{y\to+\infty} f(1/y)$$ Commented Apr 13, 2017 at 17:28
Because the left limit is $-1$ and the right is $+1$