How to prove $a^x \ll x^b$ for $b>0$ and $a>1$ with the exponential function?
My attempt:$$\lim\limits_{x\to\infty}\frac{x^b}{a^x}=\lim\limits_{x\to\infty}\frac{\exp(\log(x))^b}{\exp(\log(a))^x}$$ I don't know how to continue... any ideas, suggestions, hints? (No L'hospital!)
$$\lim_{x\to \infty} \frac {x^b}{a^x}=\lim_{x\to\infty}\frac {e^{b\cdot logx}}{e^{x\cdot loga}}=\lim_{x\to\infty}{e^{b\cdot logx-x\cdot loga}}$$ So we want to demonstrate $\lim_{x\to\infty} x(b\frac {logx}x -a)=-\infty$. It is true because $a>0$ and $lim_{x\to\infty} \frac{logx}x=0$, infact $$\frac{log x}x=log\sqrt[x]x$$ and $x^{1/x}\to 1$ as $x\to\infty$