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".   
 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.
A: You can use the definition of $x^a$ for any $a$:
$$\frac{x^a}{\mathrm e^x}= \frac{\mathrm e^{a\ln x}}{\mathrm e^x}=
\mathrm e^{a\ln x-x}.$$
Now, as $\ln x=_\infty o(x)$, we have
$$a\ln x-x=x\Bigl(a\dfrac{\ln x}x-1\Bigr)\sim_\infty -x\xrightarrow[x\to\infty]{}-\infty.$$
A: I think it is okay to assume that $a$ is finite, even though $a$ is not specified.  Thus there exists an integer $n$ such that $n\ge a$.  Let $n =\lceil a \rceil$.
So $x^n \ge x^a, \quad \textrm{whenever } x \ge 1.$
Then, as you suggest, apply L'Hospital's rule $n$ times.
$$0 \le \lim_{x \rightarrow \infty} \frac{x^a}{e^x} \le \lim_{x\rightarrow \infty} \frac{x^n}{e^x} = \lim_{x\rightarrow \infty} \frac{n x^{n-1}}{e^x} = \cdots = \lim_{x \rightarrow \infty} \frac{n!}{e^x} = 0. $$
A: If $a\gt0$, then $x=au\to\infty$ as $u\to\infty$, hence
$$\lim_{x\to\infty}{x^a\over e^x}=\lim_{u\to\infty}{(au)^a\over e^{au}}=\left(a\lim_{u\to\infty}{u\over e^u}\right)^a=(a\cdot0)^a=0^a=0$$
provided you know that $u/e^u\to0$ as $u\to\infty$.
A: You don't need to know the value of $a$, as long as it's not a negative integer the result will follow, using L'Hospital's rule we have:
$$\lim \limits_{x \to \infty}\left( \frac{x^{a}}{e^{x}} \right)=\lim \limits_{x \to \infty}\frac{a!}{e^{x}}=0$$
