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.
Prove that for each integer $m$, $ \lim_{u\to \infty} \frac{u^m}{e^u} = 0 $
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real-analysis
calculus
limits
1 Answers
1
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.