I have to find the limit of $x^x$ as $x$ approaches $0$ without derivatives.
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3so... how far have you got? – John Dvorak Nov 09 '13 at 18:49
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1How would you apply l'Hôpital here anyway? – TMM Nov 09 '13 at 18:53
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@TMM $x^x = e^{x ln x}$. From there it's obvious with l'Hopital. – John Dvorak Nov 09 '13 at 18:55
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1How rigorous do you need to be? Can you use a table of values or a graph to estimate this? – John Engbers Nov 09 '13 at 18:56
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The limit does not exist. I assume you meant limit from the right? – user7530 Nov 09 '13 at 18:56
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@user7530 noted. I made the same assumption – John Dvorak Nov 09 '13 at 18:57
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1@Jan: It's a bit of both. In any case it would be better if the OP added his own thoughts, e.g. how he would solve this with l'Hopital and where he should use something else now. – TMM Nov 09 '13 at 18:59
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1@Jan (And some people (like me) were raised without l'Hopital's rule, so using l'Hopital here surely wouldn't be the first thing that comes to my mind.) – TMM Nov 09 '13 at 19:01
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Suggestion: Try writing $x = e^{-y}$ and taking $y \to +\infty$. Then use the fact that $e^y$ grows faster than any polynomial. – Nate Eldredge Nov 09 '13 at 19:09
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2Possible duplicate http://math.stackexchange.com/questions/473535/why-does-this-limit-exist-xx/478581#478581 – clark Nov 09 '13 at 19:14
2 Answers
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We wish to find $\lim_{x\to0^{+}}x^{x}$. Notice
$$x^{x}=e^{x\ln(x)}=e^{(-1)\frac{\ln(\frac{1}{x})}{(\frac{1}{x})}}$$
so it suffices to find $\lim_{y\to\infty}\frac{\ln(y)}{y}$.
$$\frac{\ln(y)}{y}=\frac{1}{y}\int_{1}^{y}\frac{1}{t}dt.$$
For $y\ge1$ we have that $\sqrt{y}\le y$ so $\frac{1}{y}\le\frac{1}{\sqrt{y}}$ so:
$$\frac{\ln(y)}{y}\le\frac{1}{y}\int_{1}^{y}\frac{1}{\sqrt{t}}dt=\frac{1}{y}(2\sqrt{y}-2)=\frac{2}{\sqrt{y}}-\frac{2}{y}.$$
But also $\frac{\ln(y)}{y}\ge0$ for $y\ge1$. By squeeze theorem the limit is $0$. Hence,
$$\lim_{x\to0^{+}}x^{x}=1.$$
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I'm not sure if what i'm doing is right, please feel free to add comments if i was wrong: $\lim_{x \to 0} x^x = \lim_{x \to \infty} (\frac{1}{x})^\frac{1}{x} =\lim_{x \to \infty} \frac{1}{\sqrt[x]{x}}=\frac{1}{\lim_{x \to \infty}\sqrt[x]{x}}=1$
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1I don't see how finding $\lim_{x\to \infty} \sqrt[x]{x}$ is any different – Thomas Nov 09 '13 at 19:14
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