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I have to find the limit of $x^x$ as $x$ approaches $0$ without derivatives.

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    $\begingroup$ so... how far have you got? $\endgroup$ – John Dvorak Nov 9 '13 at 18:49
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    $\begingroup$ How would you apply l'Hôpital here anyway? $\endgroup$ – TMM Nov 9 '13 at 18:53
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    $\begingroup$ How rigorous do you need to be? Can you use a table of values or a graph to estimate this? $\endgroup$ – John Engbers Nov 9 '13 at 18:56
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    $\begingroup$ @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. $\endgroup$ – TMM Nov 9 '13 at 18:59
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    $\begingroup$ Possible duplicate math.stackexchange.com/questions/473535/… $\endgroup$ – clark Nov 9 '13 at 19:14
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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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    $\begingroup$ I don't see how finding $\lim_{x\to \infty} \sqrt[x]{x}$ is any different $\endgroup$ – Thomas Nov 9 '13 at 19:14
  • $\begingroup$ Also $x^x$ is not equal to $\frac{1}{x^{1/x}}$. $\endgroup$ – daulomb Nov 9 '13 at 20:06

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