Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.

Question

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.

Answer 1

Your proof is correct.

You may consider shorter proofs by using some simplifications.

Since $x\to \infty $ we assume $x>1$

$$ \frac {x^3+1}{x^2+1} \ge\frac {x^3}{2x^2}=x/2$$

Answer 2

Your proof looks ok to me.

An option:

$x^3+1=(x+1)(x^2-x+1);$

$x^2+1<x^2+x =x(x+1)$, for $x >1$.

$\dfrac{(x+1)(x^2-x+1)}{x^2+1}>$

$\dfrac{(x+1)(x^2-x+1)}{x(x+1)}=$

$x-1+1/x >x-1.$

Answer 3

Possibly correct but unreadable.

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