I am trying to find the limit as $x \rightarrow 0$ of $x^x$ using L'Hopital's rule. I have written it in exponential form: $\lim\limits_{x \rightarrow 0} e^{x \ln x}$. I do not know how to put it in indeterminate form.
2026-03-25 11:13:40.1774437220
Limits of Indeterminate Powers in Exponential Form using L'Hopital's Rule
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$$x^x=\exp{\ln x\over (1/x)}$$ Taking derivatives on top and bottom gives $$\frac{1/x}{-1/x^2}=-x$$ So the function tends to $$\lim_{x\to 0}{e^{-x}}=e^0=1$$ Also note that this should be a one-sided limit.