Help with sequence problem, expressing it as a function of a? I'm working on a problem set for a math course right now and I've come across a problem that I am having some difficulty understanding/solving. The problem is below: 

Consider the sequence: 
for different values of a. Try to generalize the value of the limit as a function of a; give a written discussion as to why your limit value or values should be true mathematically. 

I do not understand the part where I must determine some function of a. I would really appreciate some from anyone in understanding this problem. 
Thanks guys
 A: You are given $\left\{\dfrac{n^a}{e^n}\right\}$.
Let's try arbitrary values of $a$ and see if we can establish a pattern.
Let's try $a= -100, 0,10$ respectively.
$a=-100:$
$\left\{\dfrac{n^{-100}}{e^n}\right\}=\left\{\dfrac{1^{-100}}{e},\dfrac{2^{-100}}{e^2},\ldots, \dfrac{1000^{-100}}{e^{1000}},\ldots\right\}\to 0$.
$a=0:$
$\left\{\dfrac{n^{0}}{e^n}\right\}=\left\{\dfrac{1}{e^n}\right\}\to 0$.
$a=10:$
$\left\{\dfrac{n^{10}}{e^n}\right\}$=$\left\{\dfrac{1^{10}}{e},\dfrac{2^{10}}{e^2},\ldots, \dfrac{100^{10}}{e^{100}},\ldots\right\}\to 0.$
Noting that as $n$ approaches infinity, the denominator $e^n$ grows faster than the numerator $n^a$ (regardless of what $a$ you select), and ends up being much bigger than the numerator after you get to a large enough $n$, you can conclude that the limit of the sequence converges to $0$.
A: *If $a\geq0$ then $\displaystyle0\leq\frac{n^a}{e^n}\leq\frac{n^{\lceil a\rceil}}{e^n}$ how $\displaystyle\frac{n^{\lceil a\rceil}}{e^n}$ have form $\infty/\infty$, you can use L-hopital and you proof,  $\displaystyle \lim_{n \to \infty}\frac{n^{\lceil a\rceil}}{e^n}=0$.
*If $a<0$ then $\displaystyle0\leq\frac{n^a}{e^n}\leq\frac{1}{e^n}$ and how  $\displaystyle\lim_{n \to \infty}\frac{1}{e^n}=0$. 
And then for all $a\in \mathbb{R}$,  we obtain $\displaystyle \lim_{n \to \infty}\frac{n^{ a}}{e^n}=0$.
God bless
A: A simple proof,
assuming you know that
$e^x = \sum_{k=0}^{\infty} \frac {x^k}{k!}$.
Let $m$ be an integer
such that
$m \ge a$.
Then
$e^x > \frac{x^{m+1}}{(m+1)!}$
or
$\frac{x^a}{e^x}
\le \frac{x^m}{e^x}
< \frac{(m+1)!}{x}
$
and the right side goes to zero
for large enough $x$.
