Evaluating $\lim_{x\to 0} x^x$

Question

$$ x ^{x } = e^{\ln x^x } $$ $$ \lim_{x\rightarrow 0} x^x = \;? $$

I need to find the limit of x to the power of x as x approaches to 0 using l'Hopital's rule. From previous part there is a hint that I should use the first equation somehow, however I am confused how to rearrange the equation into a fraction where both nominator and denominator have limits of 0 or infinity.

Answer 1

Use these two rules:

• For every $a>0, b$: $$\ln(a^b) = b\ln(a)$$
• If the limit $$\lim_{x\to a} f(x)$$ exists and $g$ is continuous, then

$$\lim_{x\to a}(g(f(x)) = g(\lim_{x\to a} f(x))$$

Answer 2

For this limit note that we can write

\begin{equation*} \lim_{x\to 0} x^x=\lim_{x\to 0}\exp{(\log(x^x))}=\lim_{x\to 0}\exp{x\log(x)}. \end{equation*}

Factor the exponential function out of the limit due to its continuity, ie

\begin{equation*} \exp (\lim_{x\to 0}x\log(x))=\exp(\lim_{x\to 0}\frac{\log(x)}{1/x}) \end{equation*}

The limit gives $\frac{\infty}{\infty}$ so apply L'Hopital's rule:

\begin{equation*} \lim_{x\to 0}\frac{\log(x)}{1/x}=\lim_{x\to 0}\frac{\frac{d}{dx}\log(x)}{\frac{d}{dx}(\frac{1}{x})}=\lim_{x\to 0}\frac{1/x}{-1/x^2}=\lim_{x\to 0}-x. \end{equation*}

Therefore, we have

\begin{equation*} \exp(\lim_{x\to 0}-x)=\exp(0)=1. \end{equation*}