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?
2 Answers
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
$f(x) <b\dfrac{1}{x}<b(1/M)<\epsilon$.
