# Evaluating $\lim_{x\to0}{x^x}$ [duplicate]

Find the limit: $$\lim_{x\to0^+}{x^x}$$

*This question appeared in a calculus exam, and I would like to see different approaches and solutions to the problem.

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• Which approach did you use? – yiyi Sep 5 '13 at 8:20
• @yiyi My solution/approach appears below in the answers. – NightRa Sep 5 '13 at 8:21
• Perhaps you mean $x \to 0^+$ instead of $x \to 0$? – Antonio Vargas Sep 5 '13 at 8:23
• math.stackexchange.com/questions/473535/… – njguliyev Sep 5 '13 at 8:34

Let's define the limit as $L$: $$\lim_{x\to0}{x^x}=L$$ Then, we can take a logarithm: $$\ln(L)=\ln(\lim_{x\to0}x^x)=\lim_{x\to0}{x\ln(x)}$$ $$\lim_{x\to0}{x\ln(x)}=\lim_{x\to0}{\frac{\ln(x)}{\frac{1}{x}}}$$ Now we can apply Lhopital's rule: $$\ln(L)=\lim_{x\to0}{\frac{\frac{1}{x}}{-\frac{1}{x^2}}}=\lim_{x\to0}{-x}=0$$ Then, let's return to L: $$\ln(L)=0\implies L=e^0=1$$