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What is the value of this limit? $$ \lim_{x \to \infty}x^{\frac{1}{x}} $$

I have never encountered such a limit before, so any help or advice would be much appreciated.

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What's the definition of $x\mapsto x^{1/x}$? – Git Gud Jun 29 '14 at 13:10
What is it, sir? – user34304 Jun 29 '14 at 13:12
It is $e^{\frac {\log(x)}x}$, for all $x>0$. It's undefined for $x\leq 0$. – Git Gud Jun 29 '14 at 13:13
Funny. Why did two people up vote the OP's question in the comment? – Git Gud Jun 29 '14 at 13:20
Then I guess they're glad the OP asked? – G Tony Jacobs Jun 29 '14 at 13:25

Here's a start:

$\lim x^{1/x} = \lim \exp(\log(x^{1/x)})) = \lim \exp\left[\frac1x\log x\right] = \exp\left[\lim\frac1x\log x\right]$.

The limit in that last expression is a $0\cdot\infty$ form. Do you know how to handle those with L'Hôpital's Rule?

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Yeah,I can take it from here. Thanks for your help – user34304 Jun 29 '14 at 13:36
@user34304 If this answer solves your problem, please accept it. – Ruslan Jun 29 '14 at 19:31


Use the L'Hôpital's rule to find

$$\lim_{x\to\infty}\frac{\ln x}{x}$$

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Wow, you've been active! And have done quite well (no surprise!) – amWhy Jun 30 '14 at 12:58

An approach similar to G Tony Jacobs: use the continuity of logarithm (i.e. $\log \lim f(x) = \lim \log f(x)$) to log the expression to get $$ L f(x) = \frac{\log x}{x} $$ then show it converges to $0$ by L'Hospital's rule, then exponentiate back to get 1.

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