# How to prove $\lim_{x \to \infty} \dfrac{x^a}{x^b} = 0$ for all nonnegative constants $a < b$.

I'm working my way through Mathematics for Computer Science at MIT OCW, and there is a lemma in the text that I am trying to prove and I've gotten stuck.

$$\lim_{x \to \infty} \dfrac{x^a}{x^b} = 0$$ for all nonnegative constants $$a < b$$

I got this far:

\begin{align*} \ln(\lim_{x \to \infty} \dfrac{x^a}{x^b}) &= \lim_{x \to \infty} \ln \left(\dfrac{x^a}{x^b} \right) \\ &= \lim_{x \to \infty}\big( \ln(x^a) - \ln(x^b)\big) \\ &= \lim_{x \to \infty}\big( a\ln(x) - b\ln(x)\big) \\ &= \lim_{x \to \infty}a\ln(x) - \lim_{x \to \infty} b\ln(x) \\ &= a\lim_{x \to \infty}\ln(x) - b\lim_{x \to \infty}\ln(x) \\ \end{align*}

My calculus is weak, so I got stuck here and just couldn't close the deal.

• Try simplifying first. We have $\frac{x^a}{x^b} = x^{a-b} = \frac{1}{x^{b-a}}$ Commented May 20, 2016 at 0:12
• Doh. I guess my algebra sucks too. Thanks. I will try this. Commented May 20, 2016 at 0:12

$\lim_{x\to\infty} \frac{x^a}{x^b}=\lim_{x\to\infty}x^{a-b}=\lim_{x\to\infty}\frac{1}{x^{b-a}}$
Since $a<b$, we have $b-a>0$. Since the function $x^{b-a}$ is monotonically increasing and x goes to infinity, $x^{b-a}$ also goes to infinity. Therefore $\lim_{x\to\infty}\frac{1}{x^{b-a}}=0$