Proving a Polynomial Limit where x → 0 Prove
$$\lim \limits_{x\to 0} x^3+x^2+x = 0$$
Note
$|f(x)-L| = |x^3+x^2+x|$
Assume
$\ \ |x-c|<\delta \implies |x| < \delta$
$\implies |x^3+x^2+x|<\delta\cdot|x^2+x+1|$
Assume $|x| < 1 \implies -1 < x < 1 \implies 0 < x+1 < 2$
And I am not sure where to go from there, since I can't multiply the inequality by $x$ in order to get $x^2$, because $x$ could be negative or positive.

As an alternative by squeeze theorem assuming $|x|<1$
$$0\le |x^3+x^2+x|=|x||x^2+x+1|\le 3|x| \to 0$$
