# Prove that $\lim_{u\to \infty}{\frac{u^m}{e^u}}=0$ [duplicate]

I'm trying to prove that $\lim_{u\to \infty}{\frac{u^m}{e^u}}=0$ for any integer $m$. I tried using the Ratio Test For Limits to prove that the limit converges to $0$ for all real values of $m$, and therefore it must converge to $0$ for all integer values of $m$ since $\mathbb{Z} \subset \mathbb{R}$. However, using the Ratio Test only seems to complicate things. Is there a better way to show this?

Additionally, I need to prove that $\lim_{x\to 0^+}{x\log x}=0$ by setting $u=-\log x$ and using the above limit. So far, I have done the following:

$$\lim_{-\log x\to \infty}{\frac{(-\log x)^m}{e^{-\log x}}}=\lim_{x\to 0^+}x{(-\log x)^m}$$

but I don't know where to go from here. Also, I cannot use L'Hopital's rule, since I haven't covered it in lectures. Thanks in advance.

## marked as duplicate by Yanior Weg, Shailesh, Pacciu, Lee David Chung Lin, Lord Shark the UnknownApr 25 at 2:01

• Are you allowed to use L'Hopital's Rule? If so, use it repeatedly (about $m$ times:). For the limit of $x \log(x)$, introduce the new variable $z = 1/x$. Then explore the limit of $-\log(z) / z$ as $z \rightarrow \infty$. Again, L'Hopital should do it. – avs Feb 15 '17 at 22:40
• Are you supposed to know $\lim_{x\to\infty}\dfrac{\ln x}x=0$.? – Bernard Feb 15 '17 at 22:42
• I haven't covered L'Hopital, so I'm afraid I can't use it – aL_eX Feb 15 '17 at 22:47
• It's a high-school result, though. You can show it by showing first that $\ln x <2\sqrt x$ for all $x$. – Bernard Feb 15 '17 at 23:02

If you accept that $e^{u}=\sum_{n=0}^{\infty} \frac{u^n}{n!}$, then

$$\frac{u^m}{e^u}=\frac{u^m}{\sum_{n=0}^{\infty} \frac{u^n}{n!}}\le (m+1)!\frac{u^m}{u^{m+1}}\to 0$$

as $u\to\infty$.

Notice that:

$$\lim_{u\to\infty}\frac{u^m}{e^u}=m\lim_{u\to\infty}\frac{u^{m-1}}{e^u}$$

Which follows from L'Hospital's rule. Since it should be obvious that

$$\lim_{u\to\infty}\frac{u^k}{e^u}=0\forall k\le0$$

Then it follows by induction that

$$\lim_{u\to\infty}\frac{u^m}{e^u}=0\forall m\in\mathbb R$$

Hint: Use l'Hopital's rule and reduce until the numerator doesn't tend to $\infty$.

Use the Taylor series to show that there exists some positive integer $M>m$ such that,

$$e^u \geq \frac{u^M}{M!}$$

For $u \geq 0$.

This implies that for $u \geq 0$,

$$|e^{-u}| \leq |\frac{M!}{u^M}|$$

$$|e^{-u}u^m| \leq |M!u^{m-M}|=M! |\frac{1}{u^{M-m}}| \to 0$$