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Is it possible to use L'hospital rule for a function whose left hand limit and right hand limit are different.

For example in the question $\lim_{x\rightarrow0} \frac{e^{1/x}-1}{e^{1/x}+1}$ The Left hand limit is equal to -1 while the right hand limit is equal to 1. However using l'hospitals rule gives $$\lim_{x\rightarrow0} \frac{e^{1/x}-1}{e^{1/x}+1}=\lim_{x\rightarrow0} \frac{-e^{1/x}/x²}{-e^{1/x}/x²}=1$$

Why does the rule gives the value of the right hand limit and not of the left hand limit. Is there any utility in using the l'hospitals rule for a evaluating a limit that does not exist (left hand limit and right hand limit are different)

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  • $\begingroup$ @Marco false. Graph the function. $\endgroup$
    – Kamal Saleh
    Commented Dec 12, 2023 at 14:17
  • $\begingroup$ @KamalSaleh thanks i don't know what i was thinking. $\endgroup$
    – Marco
    Commented Dec 12, 2023 at 14:21
  • $\begingroup$ No problem. If you like an answer, you could click the up arrow and click the green checkmark to accept it. $\endgroup$
    – Kamal Saleh
    Commented Dec 12, 2023 at 14:21
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    $\begingroup$ Just in case: $\frac{d}{dx}(e^{1/x})=-\frac{e^{1/x}}{x^2}$, not $e^{1/x}$. $\endgroup$
    – Taladris
    Commented Dec 12, 2023 at 23:02

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The problem here is that we don't have an indeterminate form on both sides. As $e^{1/x}$ approaches $0$ from the left, it becomes zero, but it approaches infinity from the right. So for the left-hand side case, $\frac{e^{1/x}-1}{e^{1/x}+1}=\frac{0-1}{0+1}$ is not an indeterminate form, but for the right-hand side, it is. So in reality you were just calculating $$\lim_{x\rightarrow0^+} \frac{e^{1/x}-1}{e^{1/x}+1}$$And not $$\lim_{x\rightarrow0} \frac{e^{1/x}-1}{e^{1/x}+1}$$

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  • $\begingroup$ Thank you sir. I am new to limits and thus I had such a doubt. So if we apply l'hospitals rule to a limit which doesn't exist (right hand limit and left hand limit are different), we get the limit (right or left) depending on which one is indeterminate $\endgroup$
    – Vignesh
    Commented Dec 12, 2023 at 14:28
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    $\begingroup$ This type of limit isn't something you should deal with every day. But I guess that you are right. If both sides approach different indeterminate forms, we would probably get a similar story. $\endgroup$
    – Kamal Saleh
    Commented Dec 12, 2023 at 14:31

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