# Limit of $x^x$ without L'Hôpital

I was trying to calculate $\lim_{x \to 0^{+}} x^x$ without L'Hôpital's rule but could not make progress.

My best shot was to show that $\lim_{x \to 0^{+}} x\ln x = 0$ as that would imply the first limit. Can anyone help me?

• Why without l'Hospital rule? – Berci Mar 6 '16 at 1:17
• I´m studying for test where I will not be allowed to use it. – Gabriel Mar 6 '16 at 1:19
• Ah, ok. Next step might be to use $x=e^t$ then we look for $e^t\cdot t$ while $t\to -\infty$. – Berci Mar 6 '16 at 1:20

Hint: What else do you know about $\ln x$? How does its growth rate compare to $\frac1x$ or $x$?

Another hint: you can transform this to a $\lim_{y \rightarrow \infty}$ problem i.e. by setting $y=1/x$. I find this much easier to work with. Thinking about things close to $0^+$ is hard.

You will want to use some combination of finding upper / lower bounds, and technically you will apply the squeeze theorem.

• I´m trying to show that ln(x) grows asymptotically slower than x. That is, $\lim_{y \to \infty} \frac{\ln y}{y} = 0$. – Gabriel Mar 6 '16 at 1:30
• @GabrielRibeiro that's well known. L'H is one way. Consider the sequence $e, e^2, e^3, \ldots$. – djechlin Mar 6 '16 at 1:33
• made it! thanks! – Gabriel Mar 6 '16 at 1:38

To do this from scratch, here's one approach:

First note that $x\ln x<0$ as soon as $x<1$.

Next, observe that $ne^{-n}\to 0$ as $n\to \infty ,\$ because $ne^{-n}<\frac{n}{1+n+\frac{n^{2}}{2}}$.

And since $(x\ln x)'=\ln x+1,\ h(x)=x\ln x$ is decreasing on $(0,e^{-n})\$ for all integers $n\geq 1$, which means we must have $x\ln x>-ne^{-n}$ for all $x\in (0,e^{-n})$.

Putting this all together then, we can write

$-ne^{-n}<x\ln x<0$ whenever $x\in (0,e^{-n})$ and now to finish, we need only appeal to the squeeze theorem to obtain the result.

In THIS ANSWER I showed using on the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\log(x)\le x-1 \tag 1$$

for $x>0$.

Using $\log(x^{\alpha})=\alpha \log(x)$ for all $\alpha$, we have for any $\alpha >0$ and $x>0$

$$\frac{x-x^{1-\alpha}}{\alpha}\le x\log(x)\le \frac{x^{1+\alpha}-x}{\alpha}$$

whereupon choosing $\alpha <1$ and applying the squeeze theorem reveals

$$\lim_{x\to 0^+}\left(x\log(x)\right)=0$$

Finally, we have

\begin{align} \lim_{x\to 0^+}x^x&=\lim_{x\to 0^+}e^{x\log(x)}\\\\ &=e^{\lim_{x\to 0^+}\left(x\log(x)\right)}\\\\ &=1 \end{align}

TOOLS USED:

(i) The limit definition of the exponential function;

(ii) Bernoulli's Inequality;

(iii) $\log(x^{\alpha})=\alpha \log(x)$;

(iv) The Squeeze Theorem;

(v) Continuity of the exponential function;

(vi) $\lim_{x\to a}f(g(x))=f\left(\lim_{x\to a}g(x)\right)$ when $f$ is continuous.

Consider $A=x^x$. Taking logarithms $\log(A)=x\log(x)$ which goes to $0$ if $x\to 0$. So $\log(A)\to 0 \implies A \to 1$.