If $ a > 1 $, then
$$ \lim_{x \to \infty} \frac{a^x}{x} = \infty. $$
Above one can find the limit which is to be shown. Any tips? The epsilon/delta method, taylor series, definition of a derivative, etc, are not accepted as it is viewed as circular reasoning. Anyways, I really do not know how start but I guess that I should use the squeeze theorem someway down the track.
Let $a=1+\varepsilon$ with $\varepsilon>0$ then binomial shows $$a^x=(1+\varepsilon)^x>\dfrac{x(x-1)}{2}\varepsilon^2$$ this proves your assertion for $x>1$, because $x\to\infty$.
Clearly, for $x>\frac{1}{\log a}$ the function $f(x)=\frac{a^x}{\log x}$ is monotone increasing.
Let, $\lfloor x\rfloor=n$
Then, for $n>\frac{1}{\log a}$ we have,
$$\frac{a^{n+1}}{n+1}\ge\frac{a^x}{\log x}\ge\frac{a^{n}}{n}$$ Again, it's easy to see that $\frac{a^{n}}{n}$ is unbounded above.
So,$\lim _{n\rightarrow\infty}\frac{a^{n}}{n}=\infty$ Hence, by Sandwich theorem,
$\lim _{x\to\infty} \frac{a^{x}}{x}=\infty$ and done!
