# Finding limits by L'Hospital's Rule

Let $f(x) = x^{\frac{1}{x}}$ for $x>0$

Calculate $\lim_{x\to0^+} f(x)$ and $\lim_{x\to\infty} f(x)$

My attempt:

First rewrite the function to a form where we can apply L'Hospital's Rule:

$x^{\frac{1}{x}} = e^{\ln(x)^{\frac{1}{x}}} = e^{\frac{\ln(x)}{x}}$

Then take the limit of the power:

$\lim_{x\to\infty} {\frac{\ln(x)}{x}}$, by L'Hospital's Rule we have:

$\lim_{x\to\infty} {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0$

This means that:

$\lim_{x\to\infty} x^{\frac{1}{x}} = e^{\lim_{x\to\infty}{\frac{\ln(x)}{x}}} = e^0 = 1$.

For the other limit (this is where I am struggling):

Again we can rewrite the function to a form where we can apply L'Hospital's Rule:

$x^{\frac{1}{x}} = e^{\ln(x)^{\frac{1}{x}}} = e^{\frac{\ln(x)}{x}}$

Then, again, take the limit of the power:

$\lim_{x\to0^+} {\frac{\ln(x)}{x}} = \lim_{x\to0^+} {\frac{\frac{1}{x}}{1}} = \frac{\infty}{1}$

Therefore we should have:

$\lim_{x\to0^+} x^{\frac{1}{x}} = e^{\lim_{x\to0^+}{\frac{\ln(x)}{x}}} = e^\infty= \infty$.

However, by looking at the graph of the function, this is clearly not the case. What am I doing wrong?

Edit: I answered my own question below.

• In the second case the limit is $-\infty$, without l'Hospital's rule. (Hence the final limit is 0). – Kola B. Jan 20 '16 at 2:16
• Okay, I think I see what you are saying, i.e. $\lim_{x\to0^+} {\frac{\ln(x)}{x}} = -\infty$ so $e^{-\infty} = 0$ ? – patrickh Jan 20 '16 at 2:30
• Yes. And the first limit is obvious, because we know limits of $\ln x$ and $1/x$ at $0+$. – Kola B. Jan 20 '16 at 2:32

$lim_{x\to0^+} {\frac{\ln(x)}{x}} = -\infty$ so $e^{-\infty} = 0$, which is indeed the solution which we can verify by looking at the graph of the function.
• It's not just not needed. L'Hopital doesn't work unless the form is indeterminate. Since $\frac{-\infty}{+0}$ is not indeterminate, L'Hopital will potentially give a wrong answer. – Thomas Andrews Jan 20 '16 at 2:40