How to prove it by epsilon-delta proof instead of l'Hôpital's rule？I do not know how to choose delta in this case.
2026-04-04 01:01:25.1775264485
$\epsilon - \delta$ proof of $\lim_{x \rightarrow 0^+} x^x = 1$
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Since $\log$ and $\exp$ are continuous functions, that is the same as proving $$\lim_{x\to 0^+}x\log x = 0,$$ or, by replacing $x$ with $e^{-t}$, $$ \lim_{t\to +\infty} -t e^{-t} = 0 $$ that is trivial since $e^t>1+t+\frac{t^2}{2}$ for any $t>0$.