# How to prove $\lim_{x \to \infty}\frac{\log(x)}{x}=0$? [duplicate]

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.

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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 $$\lim_{y\to\infty}\dfrac{y}{e^y}=0$$