Q:

Prove that $\lim_{x\to \infty} \frac{\log(x)}{x} = 0$

Prove 
  $$\lim_{x\to \infty} \dfrac{\log(x)}{x} = 0$$
  and
  $$\lim_{x\to \infty} \dfrac{\log(x)}{x^n} = 0$$

From the definition of $\log(x)$,
$$\log(x) = \int_1^x  \dfrac{1}{t} dt$$
Since $1$ is the $\sup\{f(t) : m_{i-1} \leq t \leq m_i\}$, it follows that
$$ \int_1^x  \dfrac{1}{t} dt \leq U(f, P) < x - 1$$
So $$\log(x) < x - 1 < x \Rightarrow \dfrac{\log(x)}{x} < 1$$
Up to here, I was stuck. I'm thinking of using Squeeze theorem with 
$$\dfrac{1}{x} \leq \dfrac{\log(x)}{x} \leq g(x)$$
for some valid $g(x)$. But I couldn't think of such a $g(x)$.
Could anyone share me some ideas on how to solve this problem? 
Note L'Hospital's rule is not allowed. 

A:

Your bound $\log x<x-1$ for $\log x$ can be improved: $\int_1^x\frac1tdt<\int_1^x\frac{1}{\sqrt{t}}dt$.

