Finding $\lim_{x \to \infty} x^{1/x}$

Question

I set $\lim_{x \to \infty} x^{1/x}=L$ What I tried is this:

$$\lim_{x \to \infty} \frac 1x \ln x=\ln L$$

$$\lim_{x \to \infty} \ln e^{\large \ln \frac 1x} \ln x=\ln L$$

We substitute in $e=\lim_{x \to \infty} (1+\frac 1x)^x$

$$\lim_{x \to \infty} \ln (1+\frac 1x) \ln x=\ln L$$

$$L= \lim_{x \to \infty} (1+\frac 1x)^{\ln x}$$

This looks a lot like the definition of $e$ and from here I think I could make an argument that $L=1$ because $\ln x$ increases slower than $ax$ for any $a$ after a certain point, but I would like to have a proof that is more direct. How should I proceed from here? Thanks.

P.S. No L'hopital's rule please

Why do I keep seeing these questions with "l'Hôpital forbidden"? Is this a common trick to make exercises more challenging? — not all wrong, Jul 26 '13 at 06:41

not all wrong · Answer 1 · 2013-07-26T06:59:30.603

The most direct proof if you're allowed to assume that "$\ln$ increases slower than any $ax$" for $a>0$, so that $\ln x< ax$ for sufficiently large $x$, just notes that then

$$0< \ln x /x < ax/x=a$$

Hence the sequence ends up in $(0,a)$ for all $a>0$ and so converges to 0.

By the way, to see the above just set $u=\ln x$. Then you have $u/e^u<u/(1+u+u^2/2)\to 0$ as $x\to \infty$. This gives the result you wanted directly as well.

Vishal Gupta · Answer 2 · 2013-07-26T07:07:10.340

2

You have that $\ln L = \lim \frac{1}{x}\ln(x)$. Now use L'hospital Rule.

EDIT: To my defense, I may add that I wrote this answer before L'hospital was forbidden.

edited Jul 26 '13 at 07:07

answered Jul 26 '13 at 06:37

Vishal Gupta

6,946

I forgot to mention, no L'hopital's rule – Ovi Jul 26 '13 at 06:38

score 2 · Answer 3 · answered Jul 26 '13 at 06:38

2

You know that $\log x$ is smaller than any power, hence $\log L = 0,$ and $L=1$.

answered Jul 26 '13 at 06:38

Nathaniel Bubis

33,018

score 2 · Answer 4 · answered Jul 26 '13 at 06:42

2

Your limit is equivalent to $$ \lim_{y\rightarrow 0}\frac{1}{y^y} $$ after the substitution $x\rightarrow\frac{1}{y}$ but then $\lim_{y\rightarrow 0}y^y=1$ and you are done.

answered Jul 26 '13 at 06:42

Jon

5,457

I also forgot to mention that prior knowledge of this limit is not assumed, but +1 anyway because it's my falut – Ovi Jul 26 '13 at 06:44

Shaswata · Accepted Answer · 2013-07-26T07:22:39.957

1

Put $n^{1/n}=1+\alpha$ $$n=(1+\alpha)^{n}=1+n\alpha +\frac{n(n-1)}{2}\alpha^{2}+\cdots$$

Notice that $\alpha$ has to be positive.

It can be seen that, $$n>1+\frac{n(n-1)}{2}\alpha^{2}\\\rightarrow \frac{2}{n}>\alpha^{2}$$

However $\lim_{n\rightarrow \infty}\sqrt{\frac{2}{n}}=0$.Therefore,$\alpha=0$

$$\lim_{n\rightarrow \infty}n^{1/n}=1$$

edited Jul 26 '13 at 07:22

answered Jul 26 '13 at 06:58

Shaswata

5,068

It does work for real numbers:http://www.physicsforums.com/showthread.php?t=240990 – Shaswata Jul 26 '13 at 07:07
It does hold-Have a look at "Proof when r is any real number":http://www.trans4mind.com/personal_development/mathematics/series/binomialTheorem.htm – Shaswata Jul 26 '13 at 07:22
Ok, now I understand. – Samrat Mukhopadhyay Jul 26 '13 at 07:24

score 1 · Answer 6 · answered Jul 26 '13 at 07:13

Take $\displaystyle f(x)=\frac{\ln x}{x}$ Then $f(x)$ is a decreasing function for $\displaystyle x>e (f'(x)=\frac{1-\ln x}{x^2})$, also $f(x)>0\ \forall x\in \mathbb{R}$ and given any $\epsilon >0$, you can always choose some $x>e$ such that $f(x)<\epsilon$. Hence $$\inf_{x>e}f(x)=0$$ So $$\lim_{x\rightarrow \infty}f(x)=0$$ and hence $$\lim_{x\rightarrow \infty}x^{\frac{1}{x}}=1$$

Finding $\lim_{x \to \infty} x^{1/x}$

6 Answers6