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)
Best Answer
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 $$\lim_{x\rightarrow0} \frac{e^{1/x}-1}{e^{1/x}+1}$$