Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$ Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$.
P.S. This problem comes from my analysis book. You may use the definition of limits or invoke the Heine theorem for help. It means the proof should only use some basic properties and definition of limits rather than more complicated approaches.
 A: Applying the result
Theorem: If ${a_n}$ be a sequence such that $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}= a\,,$ then
1) if $|a|<1$, then $\lim_{n\to \infty}a_n =0 \,,$ 
2) if $ a>1$, then $\lim_{n\to \infty}|a_n| =\infty \,,$
, we have, let $b_x=\frac{x^k}{a^x}$, then
$$ \lim_{x\to \infty}\frac{b_{x+1}}{b_x}=  \lim_{x\to \infty}\frac{(x+1)^k}{a^{x+1}}\frac{a^x}{x^k} = \lim_{x\to \infty}\frac{1}{a}(1+\frac{1}{x})^k = \frac{1}{a} < 1, $$
which implies by part $(1)$ of the theorem that $\lim_{x \to \infty}b_x=0. $
A: $$x>0:$$
$$\frac{a^x}{x^k}=\sum_{n=0}^{\infty}\frac{(x\ln a)^n}{x^kn!}$$
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;=\frac{1}{x^k}+\frac{\ln a}{x^{k-1}}+\cdots+\frac{(\ln a)^k}{k!}+\frac{x(\ln a)^{k+1}}{(k+1)!}+\cdots$$
$$\;>\frac{x(\ln a)^{k+1}}{(k+1)!}$$
$$\Rightarrow0<\frac{x^k}{a^x}<\frac{1}{x}\cdot\frac{(k+1)!}{(\ln a)^{k+1}}$$
$$\text{etc.}$$
A: Hard to know what is allowed. Let $x^k=y$. We are then looking at $\frac{x}{(a^{1/k})^y}$. 
Let $a^{1/k}=b$. We are computing the simpler-looking $\lim_{y\to\infty}\frac{y}{b^y}$. 
Assume that we know that $b^y$ is increasing. Let $b=1+d$. 
Then $(1+d)^y \ge (1+d)^{\lfloor y\rfloor}$. But by the Binomial Theorem, $(1+d)^{\lfloor y\rfloor} \gt d^2\frac{\lfloor y\rfloor(\lfloor y\rfloor-1)}{2}$. This shows that $b^y$ grows sufficiently faster than $y$.  
