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$$\lim_{x \to 0}x^x = 1$$

Some sources say that this is solvable L'hopital rule, and I am unsure how I can use the rule to prove this. Can anyones show me how?

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Note that if $x\lt 0$ then the exponential makes no sense. So we really are only interested in the limit as $x\to 0^+$. –  André Nicolas Oct 14 '12 at 0:53

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up vote 3 down vote accepted

$x^x = e^{x\ln x}$, so $$\lim_{x\to 0}x^x=\lim_{x\to 0}e^{x\ln x}=e^0=1$$.

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We can rewrite $x^x=e^{x\ln x}$, now using continuity of exponentiation we know that $$\lim_{x\to 0}e^{x\ln x}=e^{\lim_{x\to 0} x\ln x}$$

Calculating $\lim\limits_{x\to0} x\ln x$ is simpler, and it is indeed $0$ (you can use L'Hospital to prove this limit), now we have: $$\lim_{x\to 0}x^x=\lim_{x\to0} e^{x\ln x}=e^0=1.$$

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