Exercise 8.4.5

Question

Show $\lim_{x \to \infty} x^n e^{-x} = 0$ for all $n = 0,1, 2, \dots$.

Ulisse Mini · Accepted Answer

Note that for a fixed $n$, $\forall K\in \mathbf{R}$,we can find $N > 0$ so that whenever $x \geq N$, $e^x x^{-n} > K$. We do this by noting
$$\frac{E(x)}{x^n} > \frac{x^{n+1}}{(n+1)! x^n} = \frac{x}{(n+1)!}$$
and setting $N = K(n+1)! $.

Now, let $\epsilon > 0$, and find $M$ so that $x \geq M$ implies $e^x x^{-n} > 1/\epsilon$. Then
$$x^n e^{-x} = \frac{1}{x^{-n}e^x} < \epsilon$$
as desired.

Exercise 8.4.5

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Exercise 8.4.5

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