# Show that $x^x$ grows faster than $b^x$ as $x \to \infty$ for $b > 1$

I have been searching for this for a while, but I can't understand it from my textbook. I am supposed to:

"Show that $x^x$ grows faster than $b^x$ as $x \to \infty$ for $b > 1$"

I can't figure out why the value of $b$ really matters when they have the same limit and exponent.

• @Itay4 Then again, $x^x$ also grows faster than $(\frac12)^x$ ... – Hagen von Eitzen May 8 '17 at 13:01
• Additionally, the function $(\frac{x}{b})^x$ is strictly increasing for $x/b > 1$. – Toby Mak May 8 '17 at 13:01
• @HagenvonEitzen $b>1$ – The Dead Legend May 8 '17 at 13:01
• How about showing that $\ln x^x$ grows faster than $\ln b^x$? – Hagen von Eitzen May 8 '17 at 13:02
• The value of $b$ matters simply because if $0<b<1$, $b^x$ decreases to $0$! – Bernard May 8 '17 at 13:04

The intuition behind this is that while both $x^x$ and $b^x$ for $b>1$ have the same limit, the first expression in increasing in both the base and the exponent, whereas the second is only increasing in the exponent. So, for any fixed $b$, $x^x$ will eventually attain a base greater than $b$ and so will grow faster in the long run.

The math behind it is to look at $$\lim_{x\to\infty}\frac{x^{x}}{b^{x}}$$ and show that goes to infinity for any $b>1$, though I'll let you fill in the details.

If $x>b$ then $\frac{x}{b}>c>1$ and when $x\to \infty$ we have

$$\left(\frac{x}{b}\right)^x>c^x\to \infty$$

For $\dfrac xb>1$, $\left(\dfrac xb\right)^x$ goes to infinity.

$b^x=o(x^x)$ when $x \longrightarrow \infty$

$\lim_{x \longrightarrow \infty} \frac{b^x}{x^x}=\lim_{x \longrightarrow \infty}e^{-x(\ln{x}-\ln{b})}=0$