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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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. – user642796 Sep 05 '13 at 08:14
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Which approach did you use? – yiyi Sep 05 '13 at 08:20
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@yiyi My solution/approach appears below in the answers. – NightRa Sep 05 '13 at 08:21
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Perhaps you mean $x \to 0^+$ instead of $x \to 0$? – Antonio Vargas Sep 05 '13 at 08:23
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@AntonioVargas Probably yes, because it is otherwise not defined. – NightRa Sep 05 '13 at 08:24
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Right, it's probably obvious. Maybe it was not worth pointing out :) – Antonio Vargas Sep 05 '13 at 08:25
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1http://math.stackexchange.com/questions/473535/why-does-this-limit-exist-xx – njguliyev Sep 05 '13 at 08:34
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@NightRa, didn't see your name on the answer. – yiyi Sep 05 '13 at 09:06
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@NightRa You shouldn't have edited your solution in an answer but inside your question, which at the moment is but a PSQ. (=poorly stated question=problem statement question). – Anne Bauval Mar 05 '24 at 11:30
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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 L'Hopital'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 $L$: $$\ln(L)=0\implies L=e^0=1$$
Anne Bauval
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NightRa
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