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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
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What is it, sir? –  user34304 Jun 29 '14 at 13:12
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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
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Funny. Why did two people up vote the OP's question in the comment? –  Git Gud Jun 29 '14 at 13:20
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Then I guess they're glad the OP asked? –  G Tony Jacobs Jun 29 '14 at 13:25

3 Answers 3

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
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@user34304 If this answer solves your problem, please accept it. –  Ruslan Jun 29 '14 at 19:31

Hint

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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