How to find $\lim \limits_{x \to \infty} x^ae^{-x} = \lim \limits_{x \to \infty}\left( \frac{x^{a}}{e^{x}}\right)$ for $a>1$? How to find the limit of this:
$$\lim \limits_{x \to \infty}\left( \frac{x^{a}}{e^{x}} \right),\quad a > 1\quad=\frac{\infty}{\infty}$$
Normally, I would apply L'Hospital's rule, and repeatedly taking the derivative of the function until the limit exist without it being indeterminate, for example if a=3, however, when a is not specified... you can't do this because you don't know the value of "a" to know how many times to take the derivative of the function.
Informally, I can understand that the limit is zero because the term $e^x$ grows at an increasing exponential rate towards infinity, and the $x^a$ term  grows at a fixed exponential rate of "a".   

Let $n$ be an integer greater than $a$. Then, by series expansion of $e^x$, we get,
$$e^x>\dfrac{x^n}{n!}$$
$$\implies 0\le\lim_{x\to\infty}\dfrac{x^a}{e^x}\le\lim_{x\to\infty}\dfrac{n!}{x^{n-a}}$$
Now apply sandwich theorem.
