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How to prove that

$\lim_{x\to 0^+} x^{x} = 1$,
$\lim_{x\to 0^+} x\ln(x) = 0$

without using L'Hopital's rule.

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Poor L'Hôpital. No one ever wants to use his rule... – David Mitra Jun 21 '13 at 16:26
Perhaps you want $x\to0^+$ rather than $x\to0$? – user1551 Jun 21 '13 at 16:27
oh yeah... thanks – udiboy1209 Jun 21 '13 at 16:29
up vote 1 down vote accepted

Another answer: when $x>0$, $x^x$ is a strictly monotonic function. So it suffices to show that $\lim_{n\to\infty}\left(\frac1n\right)^{1/n}=1$, or equivalently, $\lim_{n\to\infty}\sqrt[n]{n}=1$. You can see this thread for various proofs of this. In particular, Aryabhata's answer is easy enough to swallow.

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We have to use $x^x=\exp(x\ln x)$ as definition of exponentiation. Then it suffices to show $\lim_{x\to0^+}x\ln x= 0$. Substituing $x=\frac1{e^t}$, this becomes $\lim_{t\to\infty}\frac{-t}{e^t}=0$, which you may already know. (If not, use $e^t=e^{t/2}e^{t/2}\ge (1+t/2)(1+t/2)=1+t+\frac14t^2$, from the most useful inequality about the exponential: $e^t\ge 1+t$ for all $t\in\mathbb R$).

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could you also solve for me the limit -t/e^t = 0, i don't know how to solve it – udiboy1209 Jun 21 '13 at 16:37
@udiboy I wrote down how to get that: Using $e^t\ge 1+t+\frac14 t^2>\frac14t^2$ for $t>0$, we get $\left|\frac{-t}{e^t}\right|<\frac 4t$ for $t>0$ – Hagen von Eitzen Jun 21 '13 at 16:51
I took the liberty to change an $n$ for an $x$. – Pedro Tamaroff Jun 21 '13 at 16:53

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