How does $\left|-x^{n}{e^{-x}}\right|_{0}^{\infty}$ evaluate to zero? Is there a different explanation?

Question

Gamma function, $\Gamma(n)=\int_{0}^{\infty}{e^{-x}}{x^{n-1}}{dx}\quad(n>0)$

$$\Gamma(n+1)=\int_{0}^{\infty}{e^{-x}}{x^{n}}{dx} = \left|-x^{n}{e^{-x}}\right|_{0}^{\infty}+n\Gamma(n)$$

$$\begin{align} A &= \left|-x^{n}{e^{-x}}\right|_{0}^{\infty} \\ &= \left|\frac{-x^{n}}{e^{x}}\right|_{0}^{\infty} \\ &=-\lim_{x \to \infty}\frac{x^{n}}{e^{x}}+0 \\ &=-\lim_{x \to \infty}\frac{nx^{n-1}}{e^{x}}, \qquad \text{using L' Hospitals Rule} \\ &=-\lim_{x \to \infty}\frac{n(n-1)x^{n-2}}{e^{x}}, \qquad \text{using L' Hospitals Rule} \\ &\quad \vdots \\ &=-\lim_{x \to \infty}\frac{n*(n-1)...*1*(x^{0})}{e^{x}}, \qquad \text{using L' Hospitals Rule} \\ &= 0 \end{align}$$

Answer 1

You can make your argument much simpler with the following observation:

$e^x = \sum_{k = 0}^\infty \frac{x^k}{k!}$, so certainly $e^x \ge x^{n+2}/(n+2)!$

Then for $x > (n + 2)!$, $e^x > x^{n+1}$, so $x^ne^{-x} < 1/x$.

In short, "exponentials beat polynomials". Always, eventually. So as soon as you get $x^n/e^x$, you know it goes to 0 as $x \to \infty$.

(Notice that the argument above is a huge overestimate for when this will happen!)

Answer 2

$$\left.\frac{}{}-x^ne^{-x}\right|_0^\infty:=\lim_{b\to\infty}-b^ne^{-b}+0^ne^{-0}=-\lim_{b\to\infty}\frac{b^n}{e^b}\stackrel{\text{l'Hospital, for ex.}}=0$$