Proving a Polynomial Limit where x → 0

Question

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.

Answer 1

You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$

Note that for $|x|\lt 1$ you have $|x^2+x+1|\le |x^2|+|x|+1\lt 3$

So if $\delta\lt 1$ you have $|x^3+x^2+x|\lt 3\delta$ and you can make this as small as you like.

Answer 2

Hint: Use the triangle inequality $|a+b|\leq |a|+|b|$. You can conclude $|x^2+x+1|\leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.

Answer 3

Take $\varepsilon>0$ and pick $\delta=\min\left\{1,\frac\varepsilon3\right\}$. Then, $|x|<\delta\implies|x|<1\implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<\frac\varepsilon3\implies|x^3+x^2+x|\leqslant|x|^3+|x|^2+|x|<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.$$

Answer 4

As an alternative by squeeze theorem assuming $|x|<1$

$$0\le |x^3+x^2+x|=|x||x^2+x+1|\le 3|x| \to 0$$