# Thread: What is the limit of x^(1/x) as x approaches infinity?

1. ## What is the limit of x^(1/x) as x approaches infinity?

What is the limit of x^(1/x) as x approaches infinity?

There is an explanation in the last box at the very end of this page: Pauls Online Notes : Calculus I - L'Hospital's Rule and Indeterminate Forms

But I don't understand why he set y=x^(1/x). Where did the y come from? And how did he arrive at ln x / x ?

This is the only step I'm having trouble with but it appears to be pretty basic. I'm either having a huge mind fart or I'm overlooking something.

2. ## Re: What is the limit of x^(1/x) as x approaches infinity?

Originally Posted by TWN
What is the limit of x^(1/x) as x approaches infinity?
$\displaystyle \lim_{x\to\infty}x^{1/x}$

Our variable, $\displaystyle x$, is in the exponent. That's going to be tricky to deal with. One general strategy in this situation is to use logarithms to rewrite the expression so that the variable is no longer in the exponent (recall that $\displaystyle \log a^x=x\log a,\ a>0$).

Okay, so let's assume that the limit does exist, and that $\displaystyle x^{1/x}$ approaches some finite number. Let's call this number $\displaystyle y$. Then we have

$\displaystyle y = \lim_{x\to\infty}x^{1/x}$.

Taking the natural logarithm of both sides gives us

$\displaystyle \ln y = \ln\left(\lim_{x\to\infty}x^{1/x}\right)$.

And since $\displaystyle \ln x$ is continuous (for positive $\displaystyle x$), we are allowed to bring it inside:

$\displaystyle \ln y = \lim_{x\to\infty}\ln\left(x^{1/x}\right)$.

$\displaystyle \Rightarrow\ln y = \lim_{x\to\infty}\left[\left(\frac1x\right)\ln x\right]$ (from the above property of logarithms)

$\displaystyle \Rightarrow\ln y = \lim_{x\to\infty}\frac{\ln x}x$.

Now our limit produces the indeterminate form $\displaystyle \infty/\infty$, so we may apply L'Hôpital's rule. Then we get

$\displaystyle \ln y = \lim_{x\to\infty}\frac{1/x}1=\lim_{x\to\infty}\frac1x$

$\displaystyle \Rightarrow\ln y = 0$

$\displaystyle \Rightarrow y = e^0 = 1$.

We have therefore demonstrated that $\displaystyle \lim_{x\to\infty}x^{1/x} = 1$.