# $\epsilon - \delta$-proof of $\lim_{x\to+\infty}\frac{x}{a^x}=0$

Is there a way of proving the "standard limit" $$\lim_{x\to+\infty}\frac{x}{a^x}=0, \text{if } a>1$$ with an epsilon-delta proof? I have a proof in my course literature which is stated before the formal definition of a limit, so it has a completely other approach. I'd like to see it in the context of the rigorous definitions. The general idea of proving it this, other way, would be to do some scratch-work to find an $$\omega$$ such that $$x>\omega$$ implies $$\frac{x}{a^x}<\epsilon$$ or $$\frac{a^x}{x}>\epsilon.$$ The latter might use that $$a^x>x$$ for large enough $$x$$, but I'm unsure how to go about finding a sufficient $$\omega$$-value. To solve $$\frac{x}{a^x}=\epsilon \text{ or } \frac{a^x}{x}=\epsilon$$ for $$x$$ should give a hint of this value, but how does one do that?

You can assume $$x>2$$. Using $$x<\lfloor x\rfloor+1$$, and $$a^x\geq a^{\lfloor x\rfloor}$$, we get $$\frac{x}{a^x}< \frac{\lfloor x\rfloor+1}{a^{\lfloor x\rfloor+1}}\cdot a=\frac{n}{a^n}\cdot a$$ where $$n=\lfloor x\rfloor+1>2$$. Note $$n$$ is a natural number. Since $$a>1$$, we can write $$a=1+q$$ where $$q>0$$. Considering the binomial expansion of $$a^n=(1+q)^n$$, we have $$(1+q)^n>\frac{n(n-1)}{2}q^2$$ Hence, $$\frac{n}{a^n}\cdot a<\frac{2a}{(n-1)q^2}<\frac{2a}{(x-1)q^2}$$ Then, for a given $$\varepsilon>0$$, you can find $$v$$ such than when $$x>v$$, $$\frac{2a}{(x-1)q^2}<\varepsilon$$ To incorporate the $$x>2$$ assumption, you can choose $$\omega=\max(2,v)$$.

Option:

$$a^x=e^{x\log a}$$, where $$\log a>0$$.

$$f(x):= e^{x\log a}>$$

$$1+x\log a + (x^2(\log a)^2)/2!.$$

$$\dfrac{x}{a^x} =\dfrac{x}{e^{x \log a}}< \dfrac{2x}{x^2(\log a)^2}=$$

$$\dfrac{2}{x(\log a)^2}= b(1/x)$$, where

$$b=2/(\log a)^2.$$

Let $$\epsilon >0$$ be given.

Let $$M>0$$, real, be such that $$M >(b/\epsilon).$$

Then $$x >M$$ implies

$$f(x) .