limits - Applying L'Hôpital's rule infinitely

I tried to prove that $\int\limits_0^\infty t^{x-1} e^{-t} \, \mathrm{d}t$ satisfies the functional equation of the gamma function $\Gamma(x+1)=x\Gamma(x)$, so I partially integrated $\Gamma(x+1)$, yielding $\left[-e^{-t}\,t^x\right]_0^\infty+x \Gamma(x)$.

It is obvious to me that

$$\lim_{a\to\infty} \frac{a^x}{e^a}$$ is zero for any finite $a$ due to the fact that the exponential grows faster than any polynomial. You could show this quite easily by using L'Hôpital's rule $a$ times. I can imagine, however, that this is not true in the non-finite case. Can I apply L'Hôpital in this way, what would I have to show in order to do so, and if I cannot, please give me a hint how to obtain the desired result in a different way.

Answer

I'm not exactly sure what is meant by the "non-finite" case, but if you are trying to establish the limit without using L'Hospital's rule then the following argument is applicable.

We are considering the limit as $a \rightarrow \infty$. For $a > 1$ there is a positive integer $n$ such that $a^x < a^n$. Using the Taylor series for $e^a$ we get

$$\frac{e^a}{a^n} = \sum_{k=0}^{\infty}\frac{a^k}{k!a^n} > \frac{a}{(n+1)!}$$

and

$$\frac{a^x}{e^a} < \frac{a^n}{e^a}<\frac{(n+1)!}{a}$$

Whence

$$\lim_{a \rightarrow \infty} \frac{a^x}{e^a}= \lim_{a \rightarrow \infty} \frac{(n+1)!}{a}= 0$$