How to prove $\lim_{x \to \infty}\frac{\log(x)}{x}=0$? I will soon make a math exame where one can't use L'Hôpital's rule, integrals concept neither the formal limit definition. The most I can use is the derivative definition and the algebraic ways to solve limits. My thought was: Let $\displaystyle \lim_{x \to \infty} \frac{\log(x)}{x}=y$, so $\displaystyle \lim_{x \to \infty}\log(x^{\frac{1}{x}})=y$. Then, $$\displaystyle \lim_{x \to \infty}e^{\log(x^{\frac{1}{x}})}=e^y \Leftrightarrow$$ $$\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}}=e^y $$ I ended with an indetermination. How can I proof the $\displaystyle \lim_{x \to \infty}\frac{\log(x)}{x}=0$ with the restritions and as formal it can get? Thanks.

Write $\log(x)=y$ then $x=e^y$ and as $x\to\infty$, $y\to\infty$.

You want to find $$\lim_{y\to\infty}\dfrac{y}{e^y}$$

But $e^y\geq 1+y+y^2/2!$ for any $y\in\mathbb R$ hence $$0\leq\lim_{y\to\infty}\dfrac{y}{e^y}\leq\lim_{y\to\infty}\dfrac{y}{1+y+y^2/2!}=0$$ so $$\lim_{y\to\infty}\dfrac{y}{e^y}=0$$