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

Question

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 08 '17 at 13:01
Additionally, the function $(\frac{x}{b})^x$ is strictly increasing for $x/b > 1$. — Toby Mak, May 08 '17 at 13:01
How about showing that $\ln x^x$ grows faster than $\ln b^x$? — Hagen von Eitzen, May 08 '17 at 13:02
Note that for all $x>b>1$, $x^x>b^x>1$. We are able to assume $x>b$, because $b$ is fixed and $x\rightarrow\infty$. — SaeidAli, May 08 '17 at 13:03
The value of $b$ matters simply because if $0<b<1$, $b^x$ decreases to $0$! — Bernard, May 08 '17 at 13:04
@Bernard answers this question. Also, the title is badly chosen. What does this have to do with l'Hospital's rule? — mickep, May 08 '17 at 13:08
@Bernard: so the behavior is the same for $b\ge0$ as for $b>1$ and the latter inequality is optional. — , May 08 '17 at 13:12
@mickep you are right, we have just been going over it in class and I mindlessly just titled it that. — Jeff, May 08 '17 at 13:14
@Jeff for future reference, it is possible to [edit] questions (or link below the post.) I changed the title. — quid, May 08 '17 at 16:18

score 9 · Answer 1 · answered May 08 '17 at 13:07

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.

score 2 · Answer 2 · answered May 08 '17 at 13:06

2

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

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

answered May 08 '17 at 13:06

Arnaldo

21,342

score 2 · Answer 3 · 2017-05-08T14:42:14.443

2

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

edited May 08 '17 at 14:42

answered May 08 '17 at 13:14

score 1 · Answer 4 · answered May 08 '17 at 13:09

1

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

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

answered May 08 '17 at 13:09

Marios Gretsas

21,906

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

4 Answers4