-3

I'm unsure how to show that for each integer $m$, $ \lim_{u\to \infty} \frac{u^m}{e^u} = 0 $. Looking at the solutions it starts with $e^u$ $>$ $\frac{u^{m+1}}{(m+1)!}$ but not sure how this is a logical step.

AnoUser1
  • 169
  • 2
  • 9
  • Hint: $u^m/e^u=(u/e^{u/m})^m=m^m((u/m)/e^{u/m})^m$ – egreg Apr 24 '19 at 16:20
  • Are you saying you don't understand why it's true that $e^u \gt \frac{u^{m+1}}{(m+1)!}$? Or are you saying you don't know how to use that observation? – Robert Shore Apr 24 '19 at 16:22
  • Yeah. I think I'm just forgetting a rule – AnoUser1 Apr 24 '19 at 16:23
  • Also: https://math.stackexchange.com/q/1487289/42969, https://math.stackexchange.com/q/1908942/42969 – Martin R Apr 24 '19 at 17:16
  • I disagree with the close votes. Check out the comments to the answer below. The poster isn't just looking for an answer. He or she is clearly trying to understand what's going on. – Robert Shore Apr 24 '19 at 17:20
  • @RobertShore: OP wonders about $e^u > \frac{u^{m+1}}{(m+1)!}$, and that is covered e.g. in these answers https://math.stackexchange.com/a/2146297/42969, https://math.stackexchange.com/a/1487382/42969 to the possible duplicates. – Your approach is covered e.g. here: https://math.stackexchange.com/a/2146298/42969. – Martin R Apr 24 '19 at 17:25

1 Answers1

0

By L'Hopital's Rule (taking derivatives with respect to $u$),

$$\lim_{u \to \infty} \frac{u}{e^u} = \lim_{u \to \infty} \frac{1}{e^u}=0$$

Now work by induction. Assume we know the result for $m$. Then by L'Hopital's Rule,

$$\lim_{u \to \infty} \frac{u^{m+1}}{e^u}=\lim_{u \to \infty} \frac{(m+1)u^m}{e^u}=(m+1)0=0.$$

Where the penultimate step uses our inductive hypothesis.

Robert Shore
  • 23,332
  • How do you get this? $$\lim_{u \to \infty} \frac{u^{m+1}}{e^u}=\lim_{u \to \infty} \frac{(m+1)u^m}{e^u}$$. And also, I think that the first line is then used for the next step after whats stated above, but you used just u, and not $u^m$ on the first line, so how does this help? – AnoUser1 Apr 24 '19 at 16:43
  • Do you know L'Hopital's Rule? When you have a limit in the form $\frac{\infty}{\infty}$, you can take the derivative of the numerator and denominator and the limit is the same. And our inductive hypothesis is that the statement is true for $m$ (not just for $m=1$), and based on that assumption you're trying to prove that the statement is also true for $m+1$. – Robert Shore Apr 24 '19 at 17:18