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I have to prove that $$\lim_{x \to \infty} \frac{x^{a}}{b^{x}} $$ equals zero for all $a$ of $\mathbb{N}$ and $|b|>1$

The suggestion is to start by finding $$\lim_{x \to \infty} \frac{x^{1}}{b^{x}} $$ which I have done. But what I get is $\frac{x}{b^{x}}$ and since the limit of $b^{x}$ is defined as $0$ I don't know how to continue...

Hosein Rahnama
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Ellis V.
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2 Answers2

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Since $\lim_{x\to\infty}x^a=\infty$ and $\lim_{x\to\infty}b^x=\infty$, you can to use L'Hospital's Rule:

\begin{align} \lim_{x\to\infty}\frac{x^a}{b^x} = \lim_{x\to\infty}\frac{d(x^a)/dx}{d(b^x)/dx} = \lim_{x\to\infty}\frac{ax^{a-1}}{\ln(b) b^x} \end{align}

Notice, if $a=1$ then the limit in the numerator is just $1$ while the denominator still goes to infinity, so that the limit of the ratio goes to zero. To complete the proof just notice that for any $a \in \mathbb{N}$ you can just repeat the above step $a$ times until you obtain the same result.

Edit: The above proof work for $b>1$. For the case in which $b<-1$ notice that

$$ \frac{d(b^x)}{dx} = (\ln(-b)+\pi i)b^x$$

the limit of which is $-\infty$. So an analogous argument works for that case.

mzp
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If you have done $$\lim_{x\to \infty} \frac{x}{b^x} = 0, $$ for all $b >1$, then

$$\lim_{x\to \infty} \frac{x^n}{b^x} = \lim_{x\to \infty} \left(\frac{x}{(\sqrt[n]b)^x}\right)^n = \left( \lim_{x\to \infty}\frac{x}{(\sqrt[n]b)^x}\right)^n =0$$ as $\sqrt[n] b$ is still $>1$. Not sure what's going on when $b<-1$.