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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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  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – user642796 Sep 5 '13 at 8:14
  • $\begingroup$ Which approach did you use? $\endgroup$ – yiyi Sep 5 '13 at 8:20
  • $\begingroup$ @yiyi My solution/approach appears below in the answers. $\endgroup$ – NightRa Sep 5 '13 at 8:21
  • $\begingroup$ Perhaps you mean $x \to 0^+$ instead of $x \to 0$? $\endgroup$ – Antonio Vargas Sep 5 '13 at 8:23
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    $\begingroup$ math.stackexchange.com/questions/473535/… $\endgroup$ – njguliyev Sep 5 '13 at 8:34
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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: $$\ln(L)=0\implies L=e^0=1$$

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