Evaluate $$\lim_{x \to 0^+} x^{x^x}$$
I have assumed $$L=\lim_{h \to 0}h^{(h^h)}$$ taking $x=0+h$ where $h$ is a very small positive real number.
Now taking natural Log on both sides we get
$$\ln L=\lim_{h \to 0}h^h \ln h$$
How to proceed further
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$$\lim_{x \to 0}x^x =1$$
Hence
$$\log \mathrm L=\lim_{x \to 0}x^x \log x=\lim_{x \to 0}1 \cdot \log x=-\infty$$
$$\implies \mathrm L=0$$
If you want to know why is $\displaystyle\lim_{x \to 0}x^x =1$, rewrite it as $\displaystyle\lim_{x \to 0}{e^{x \ln x}}$ and for that limit, See this.
Jaideep Khare
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sorry. But, why $\lim_{x\to 0}x^x=1$? – Julio enrique godoy muñoz May 20 '17 at 03:48
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that is more clear for me, thanks. – Julio enrique godoy muñoz May 20 '17 at 03:56
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@Julioenriquegodoymuñoz Welcome, glad I could help you! :-) – Jaideep Khare May 20 '17 at 04:17
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@JaideepKhare Sorry, I have a question $$\log L=\lim_{x \to 0}x^x \log x$$ after substituting $$x^{x}=1$$ we have to deal with logarithm $$\lim_{x\to0}=1 \cdot \log x=-\infty$$ but why the answer is not $$-\infty$$? And why we can write so $$\log L=\lim_{x \to 0}x^x \log x$$ – Ilya Telefus May 20 '17 at 06:32
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I ca understand why $$ \lim_{x\to0}x^{x}=1$$ but here we have $$ \lim_{x\to0}x^{x^{x}}$$ is the result the same? – Ilya Telefus May 20 '17 at 06:52
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@IlyaTelefus For your first comment, we have got $\log L =-\infty \implies L=e^{-\infty}$.Hence the limit is $$L=\frac{1}{e^{\infty}}=0$$ I don't understand what do you want to say in your second comment. – Jaideep Khare May 20 '17 at 07:27