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

Question

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.

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

Pedro · Accepted Answer · 2013-05-23T15:04:48.653

5

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.

edited May 23 '13 at 15:04

answered May 23 '13 at 14:53

Pedro

122,002

@ThomasAndrews You want $k\to\infty$; yes? Talking about dummy variables should prove useful, too, yes. – Pedro 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 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 May 23 '13 at 15:23

score 1 · Answer 2 · answered May 23 '13 at 14:48

1

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

answered May 23 '13 at 14:48

ronno

11,390

score 1 · Answer 3 · answered May 23 '13 at 15:06

1

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 $$1<n^{\frac1n}<1+\frac{\sqrt2}{\sqrt{n-1}}\to 1$$

answered May 23 '13 at 15:06

Hagen von Eitzen

374,180

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

3 Answers3

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