Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove that

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

without using L'Hopital's rule.

share|improve this question
9  
Poor L'Hôpital. No one ever wants to use his rule... –  David Mitra Jun 21 '13 at 16:26
1  
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

2 Answers 2

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.

share|improve this answer

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 $t\in\mathbb R$).

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.