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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  • $\begingroup$ 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
  • 1
    $\begingroup$ math.stackexchange.com/questions/473535/… $\endgroup$ – 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: $$\ln(L)=0\implies L=e^0=1$$

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