Proving something about $|f(x)|$ when the lim of $f(x)/x^2$ is known I've been trying to crack this issue for 2 days and I got pretty much nothing
Given that $f$ is a continuous function 
and the following limits exists and are finite:
$$
(1) \lim_{x\to\infty}\frac{f(x)}{x^2}             (2) \lim_{x\to-\infty}\frac{f(x)}{x^2}
$$
I need to prove that there are constants $a$ and $b$ such that for every $x\in\Bbb R$ the following is true
$$
|f(x)|\le ax^2+b
$$
I was thinking to split the proof to 3 parts $(\infty, M_1)$, $[M_1,M_2]$, $[M_2,\infty)$
Thank you very much

We can indeed divide it up in that manner. Let $L_1$ be the limit at $-\infty$ and $L_2$ the limit at $+\infty$. By definition, we have some $M_1,M_2$ such that
$$x<M_1\implies \left|\frac{f(x)}{x^2}-L_1\right|<1\quad \text{and}\quad x>M_2 \implies \left|\frac{f(x)}{x^2}-L_2\right|<1$$
and thus
$$x<M_1\implies |f(x)|<(|L_1|+1)x^2\quad \text{and}\quad x>M_2 \implies |f(x)|<(|L_2|+1)x^2$$
so if we let $a=\max(|L_1|+1,|L_2|+1)$ and $b=\sup_{x\in [M_1,M_2]}|f(x)|$ which exists by continuity, $|f(x)|\leq ax^2+b$.
