Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition. Problem
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.
Note:
The problem asks us to prove that, no matter $x \to +\infty$ or $x \to -\infty$, the limit is $\infty$,which may be $+\infty$ or $-\infty.$
Proof
$\forall M>0$,$\exists X=\max(1,M+1)>0, \forall|x|>X$：
\begin{align*}
\left|\frac{x^3+1}{x^2+1}\right|&=\left|x-\frac{x-1}{x^2+1}\right|\\&\geq |x|-\left|\frac{x-1}{x^2+1}\right|\\&\geq |x|-\frac{|x|+1}{x^2+1}\\&\geq |x|-\frac{x^2+1}{x^2+1}\\&=|x|-1\\&>X-1\\&\geq M.
\end{align*}
Please verify the proof above.

Possibly correct but unreadable.
Consider
$$
\frac{x^3+1}{x^2+1}=x-\frac{x-1}{x^2+1}>x
$$
whenever $x>1$.
