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

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 $\lim_{x\to 0^+} f(x)=1$.


Because the left limit is $-1$ and the right is $+1$

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Limits

Calculus

Real Analysis

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