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.

The question of how to find $\displaystyle\lim_{a\to\infty}\frac{a^x}{e^a}$ or limits similar to it seems to come up often here.
And L'Hopital's rule, when it finds the answer, gives little or no insight.
Every time $a$ increases by $1$, the fraction $\dfrac{a^x}{e^a}$ is multiplied by $\dfrac{(a+1)^x/a^x}{e}<\dfrac{2}{e}$ if $a>\text{something}$.
Thus the limit is less than
$$
\text{some number} \cdot \underbrace{\frac{2}{e}\cdots\cdots\cdots\frac{2}{e}}_{\text{}}
$$
no matter how many factors appear over the $\underbrace{\text{underbrace}}$.  That should tell you what the limit is.
