Compute $\lim_{x \to \infty} \frac{e^x}{x^a}$ How can I compute $\lim_{x \to \infty} \frac{e^x}{x^a}$ for some $a \in \mathbb R$ with $x^a := e^{a \log(x)}$?
I want to use only the basic properties of limits, i.e. the linearity, multiplicativity, monotonicity and the Sandwich property (no L'Hospital).
Can you give me a hint?
 A: you may find the following line of thought helpful.
choose $x_0$ so that $e^{x_0}\gt 2^a$ and write $f(x)$ for $e^x$ and $g(x)=x^a$, so that 
$$
h(x) = \frac{e^x}{x^a} = \frac{f(x)}{g(x)}
$$
then 
$$
h(2x_0)= \frac{f(x_0)^2}{2^a g(x_0)} = Kh(x_0)
$$
where $K=\frac{f(x_0)}{2^a} \gt 1$
and then we have
$$
h(2^nx_0) = K^nh(x_0)
$$
A: another method to proof it is with the definition of $e^x$  as $e^x=\sum_{n=0}^{\infty}\frac{z^n}{n!}$
Take $n \in \mathbb{N}$ such that $a<n$ than it follows that 
$$ \frac{e^x}{x^{a}} > \frac{e^x}{x^n} = \frac{1}{x^n} \sum_{i=0}^{\infty}\frac{x^{i}}{i !} > \frac{1}{x^n}\frac{1}{(n+1)!}x^{n+1}=\frac{x}{(n+1)!} \longrightarrow \infty $$ for $x \longrightarrow\infty$.
A: Here is an entirely elementary proof. First we have that if $a>1$ then 
$$n+1\leq a^n$$ for large enough $n$. This follows from the binomial theorem, 
$$(a-1)^2\frac{n(n-1)}{2} < (1+(a-1))^n$$
now chose $n$ large enough so that 
$$n+1 < (a-1)^2\frac{n(n-1)}{2}$$ which is clearly possible.
Next, $$x < a^x$$ for large enough $x$, for if $n\leq x < n+1$ then 
$$x <n+1 < a^n <a^{x} $$
Now it follows that 
$$x^k < a^x$$ for all large $x$. Just let $a=b^k$, then 
$$x < b^x$$
and taking the power $k$, 
$$x^k < a^x.$$
It follows that given $k$, 
$$x^{k+1} < a^x$$ and so
$$x<\frac{a^x}{x^k}$$ and the limit is infinity.
