<s>[INST] Showing that $\displaystyle \lim_{x \to{+}\infty}{ax^{3}+bx^{2}+cx+d} = {+}\infty$, $a>0$. I need some help proving that:
$\displaystyle\lim_{x \to{+}\infty}{ax^{3}+bx^{2}+cx+d} = {+}\infty$, with $a,b,c,d$  real numbers, $a>0$.
Its the first time i work this kind of limits and a I feel a little lost!
 [/INST] We have, for non-zero $x$,
$$ax^3+bx^2+cx+d=x^3\left(a+\frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}\right).$$ 
For large enough $x$, we have $\left|\frac{b}{x}\right|\lt \frac{a}{4}$, and $\left|\frac{c}{x^2}\right|\lt \frac{a}{4}$, and $\left|\frac{d}{x^3}\right|\lt \frac{a}{4}$. 
Thus by the Triangle Inequality, $a + \frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}\gt \frac{a}{4}$.
It follows that for large enough $x$, we have 
$$ax^3+bx^2+cx+d \gt \frac{ax^3}{4}.$$
Remark: The intuition is that when $x$ is large, the leading term $ax^3$ is "dominant," and the others are (comparatively) small. The answer formalizes this intuition. 
To be even more formal, there is an $N_1$ such that if $x\gt N_1$, then $\left|\frac{b}{x}\right|\lt \frac{a}{4}$. Similarly, there are $N_2$ and $N_3$ for the other two desired inequalities. If $N=\max(N_1,N_2,N_3)$ then all of the inequalities hold. So if $x\gt N$ then $ax^3+bx^2+cx+d\gt \frac{ax^3}{4}$. 
</s>