# Regarding $\lim_{n \to \infty} n^{\frac{1}{n}}$

Suppose $\lim_{n \to \infty} n^{\frac{1}{n}} = l \in \mathbb{R}$. The function $f(x) = x^n$ is continuous, then

$$l^n=\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right ) =\lim_{n \to \infty} n = \infty.$$

Then it follows that $l = \infty$. This is false, but I couldn't find which step is wrong. Could you please help me? Thank you for your time.

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You are right to say that you are wrong. Infinity to the power zero is what's called an indeterminate form, so it could be "anything" Tip write your limit as y = and take the natural log on both sides. Then "pull out" the 1/n up front, and then you can "Create" a denominator with factor n. Retake the limit. Do you see it now? Don't forget to realize your answer is in ln-form (you took the ln, remember?) so "e-power" it back. Show us what you get... – imranfat May 23 '13 at 14:48

You cannot swap the $n$ around, that is, inside and outside the limit. The step $$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right )$$ is wrong.

The $n$ outside the LHS is fixed, while the $n$ inside, in the RHS, varies like the other $n$. The result would be correct, if you said, say,

$$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^3=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^3 \right )$$

As Thomas pointed out, $n$ is a dummy variable, that is $$\lim_{n\to\infty}n^{1/n}=\lim_{k\to\infty}k^{1/k}=\lim_{\sigma \to\infty}\sigma^{1/\sigma}$$

so you cannot really force $n$ to "match up" in the limit.

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@ThomasAndrews You want $k\to\infty$; yes? Talking about dummy variables should prove useful, too, yes. – Pedro Tamaroff May 23 '13 at 15:02
Doh, yep. Just gonna re-post since too late to edit. Consider that $\lim_{n\to\infty} n^{1/n}=\lim_{k\to\infty} k^{1/k}$, but if you raise the latter to the nth power, what are you getting? – Thomas Andrews May 23 '13 at 15:18
@ThomasAndrews $1=1$? – Pedro Tamaroff May 23 '13 at 15:19
Well, if you do things right, yes, but if you do things the way OP did, you get a different value than raising the left side to the $n$th power. – Thomas Andrews May 23 '13 at 15:22
@ThomasAndrews Oh, sure. – Pedro Tamaroff May 23 '13 at 15:23

Why would $\lim a_n = L$ imply $\lim a_n^n = \lim L^n$?

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Just note that $\left(1+\frac{\sqrt2}{\sqrt {n-1}}\right)^n\ge1+n\frac{\sqrt 2}{\sqrt n} +\frac{n(n-1)}2\frac 2{n-1}>n$ for $n>1$, so $$1<n^{\frac1n}<1+\frac{\sqrt2}{\sqrt{n-1}}\to 1$$

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