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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?

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    $\begingroup$ Is $f$ that quotient or is that quotient the result of the limit? $\endgroup$
    – Filburt
    Commented Apr 13, 2017 at 18:04
  • $\begingroup$ 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 $\pm1$. $\endgroup$
    – Piquito
    Commented Nov 1, 2022 at 14:23

2 Answers 2

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

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  • $\begingroup$ Oh, okay I understood my mistake. But Is L'Hospital's rule, in general valid for one sided limits? $\endgroup$
    – xasthor
    Commented Apr 13, 2017 at 17:26
  • $\begingroup$ Yes, L'Hopital can be applied to one-sided limits. You can think of it via the equality $$\lim_{x\to 0^{+}} f(x)=\lim_{y\to+\infty} f(1/y)$$ $\endgroup$ Commented Apr 13, 2017 at 17:28
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Because the left limit is $-1$ and the right is $+1$

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