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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.

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Are you allowed l'Hopital's rule? If so, apply that $k$ times to get a constant over something going to infinity. – Clayton Jan 27 '13 at 4:01
@Clayton Sorry, but that rule is not allowed here. > < – ymfoi Jan 27 '13 at 4:03


$$\frac{a^x}{x^k}=\sum_{n=0}^{\infty}\frac{(x\ln a)^n}{x^kn!}$$

$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;=\frac{1}{x^k}+\frac{\ln a}{x^{k-1}}+\cdots+\frac{(\ln a)^k}{k!}+\frac{x(\ln a)^{k+1}}{(k+1)!}+\cdots$$

$$\;>\frac{x(\ln a)^{k+1}}{(k+1)!}$$

$$\Rightarrow0<\frac{x^k}{a^x}<\frac{1}{x}\cdot\frac{(k+1)!}{(\ln a)^{k+1}}$$


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You're right. But this problem expects some kind of way that use basic properties and definition of limits. – ymfoi Jan 27 '13 at 7:09
@ymfoi It would be best if you defined what you mean by basic then. Are you trying to use $\epsilon-\delta$? If so, you can apply it to the RHS in the last line. – L. F. Jan 27 '13 at 14:53

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. $

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Note that x is not necessarily a integer. You've just proved it in sequence situation. – ymfoi Jan 27 '13 at 7:04

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 $y$.

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