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Find the limit of the given function $$\lim:\lim_{x\to 0} x^x$$ Does the function has limit.

I think no but how to prove it.

Zelos Malum
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suraj kumar behera
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  • https://answers.yahoo.com/question/index?qid=20090112034044AA71yQ3 – lab bhattacharjee Dec 30 '16 at 04:58
  • I think the limit exists and it is $1$. However, as one can observer, the $x^x$ is not real when $x<0$, so probably one has to use a little bit knowledge in complex analysis to consider $e^{x \log x}$ (honestly I am not 100% sure, since I took complex analysis a while ago, feel free to correct me). In that case, with the use of L'Hôpital's rule, one can prove the limit is $1$. – k99731 Dec 30 '16 at 06:08
  • The limit exists from the other side too: if you take $y = -x$, then as y->0^+ it's $\exp(-y(\log(y) +i\pi))$ which still goes to 1. More generally, it exists along any complex direction into the origin. – spaceisdarkgreen Dec 30 '16 at 06:35
  • Look at this: http://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero – Michael Hoppe Dec 30 '16 at 06:44
  • This is not abstract algebra – Zelos Malum Dec 30 '16 at 07:01
  • and neither implicit – Duchamp Gérard H. E. Dec 30 '16 at 07:05

3 Answers3

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To be precise $$ \lim_{x\to0}x^x $$ does not exist, but the one-sided limit does $$ \begin{align} \lim_{x\to0^+}x^x &=\exp\left(\lim_{\,x\to0^+}\frac{\log(x)}{1/x}\right)\\ &=\exp\left(\lim_{\,x\to0^+}\frac{1/x}{-1/x^2}\right)\\[3pt] &=\exp(0)\\[9pt] &=1 \end{align} $$

robjohn
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If you consider $$y=x^x\implies \log(y)=x\log(x)\implies\lim_{x\to 0} \log(y)=0\implies \lim_{x\to 0}y=1$$

You could also, as k99731 commented, use $y=e^{x\log(x)}$ and use Taylor series to get $$y=\sum_{n=0}^\infty \frac{x^n\log^n(x)}{n!}=1+\sum_{n=1}^\infty \frac{x^n\log^n(x)}{n!}$$ remembering that $\lim_{x\to 0} x\log(x)=0$

Claude Leibovici
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If $y = 1/x$, then $\lim_{x \to 0^+} x^x =\lim_{y \to \infty} (1/y)^{1/y} =\lim_{y \to \infty} \dfrac1{y^{1/y}} $

But, as is well known, and has been proved here often, $\lim_{y \to \infty} y^{1/y} =1 $ so that $\lim_{x \to 0^+} x^x =1 $.

marty cohen
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