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