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.