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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 $x\to 0^+$.2012-10-14

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$x^x = e^{x\ln x}$, 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: $\lim_{x\to 0}x^x=\lim_{x\to0} e^{x\ln x}=e^0=1.$