I'm trying to show that if $f$ is a Lipschitz function then $\lim_{x \to +\infty} \frac{f(x)}{x^n} = 0 \forall n > 1 $.
So far I proved that if $f$ is a Lipschitz function $\lim_{x \to +\infty} \frac{f(x)}{x^n} \leq 0 \forall n > 1 $:
$f$ is Lipschitz so $\exists C \in \mathbb{R}: |f(x) - f(a)| \leq |x-a| \implies f(x) -f(a) \leq C|x| + C|a| \implies f(x) \leq C|x|+cte $.
Taking the limit, I have $\lim_{x \to +\infty} f(x) \leq \lim_{x \to +\infty} C|x| \implies \lim_{x \to +\infty} \frac{f(x)}{x^n} \leq \lim_{x \to +\infty} \frac{C|x|}{x^n} = 0 \forall n > 1$.
I don't know how to prove the other inequality. Any hint is appreciated!
Thanks