Find the growth rate of $\ln(x)$ Let $p>0$. Show that
$$\lim_{x\to \infty} \frac{\ln(x)}{x^p}=0$$
My thoughts:
So, this is about growth rate of $\ln(x)$ vs. $x^p$, L'Hopital not allowed, only playing by the limits. How can I estimate these functions?
 A: 
PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1 }\tag 1$$
for $x>0$


Note that for any number $a$, $a\log(x)=\log(x^a)$. Using $(1)$ (replacing $x$ with $x^a$) reveals that for $a>0$
$$\frac{x^{a-p}-x^{-p}}{ax^a}\le \frac{\log(x)}{x^p}\le \frac{x^{a-p}-x^{-p}}{a}$$
For $0<a<p$ we can apply the squeeze theorem to obtain the coveted limit

$$\lim_{x\to \infty}\frac{\log(x)}{x^p}=0$$

A: $$\lim_{x\to\infty}\frac{\log(x)}{x^p}=\lim_{y\to\infty}\frac{\log(y^{1/p})}{y}=\frac1p\lim_{y\to\infty}\frac{\log(y)}{y}$$ $$=\frac1p\lim_{z\to\infty}\frac{z}{e^z}$$
And the last limit is clearly zero. If you are wondering about the sustitutions I made, they're just $y=x^p$ and $z=\log(y)$.
How do you know that the last limit goes to zero? Make $f(z)=z/e^z$. Note that $f(1)=1/e<1/2$. Now, you can see that $$f(z+1)=\frac{z+1}{e^{z+1}}=\frac{f(z)}{e} +\frac1{e^{z+1}}<\frac2e\cdot f(z)$$ and as the decreasing rate is strictly lower that 1, we have that for all $n\in\mathbb{N},\ n>1$: $$0\leq f(n) \leq \big(\frac2e\big)^{n-1}f(1)$$ So by compression, as $n\to\infty$ you get that the limit is zero. It is sufficient to analyze the values that $f$ has in $\mathbb{N}$ because for all $z>1$ you can see that $f'(z)<0$ by just taking the first derivative.
