I don't understand what I'm doing wrong: Prove that $f(x) = x^3$ is continuous at $x = −2$.

Question

Prove that $$f(x) = x^3$$ is continuous at $x = −2$.

For this problem, please prove it by imitating the delta -epsilon approach of C1.8.

This are my steps with my solution

$$|(x+2)(x^2-2x+4)|<ε$$ $$x^2-2x+4=((x+2)-2)^2-2((x+2)-2)+4$$ $$=(x+2)^2-6(x+2)+12$$ $$=|x+2|<1$$ $$|x^2-2x+4|<1+6+12=19$$ $$|x+2|<δ$$ $$|(x+2)(x^2-2x+4)|<(ε/19)×19=ε$$

Answer 1

Indeed,$$x^3-(-2)=(x+2)(x^2-2x+4).$$Now, note that, if $|x+2|<1$, then $-3<x<-1$ and so:

• $x^2<9$;
• $|-2x|=2|x|<6$

and so $|x^2-2x+4|\leqslant x^2+|2x|+4<19$. So, if $|x+2|<\min\left\{\frac\varepsilon{19},1\right\}$, you have$$\bigl|x^3-(-2)\bigr|\leqslant\frac\varepsilon{19}\times19=\varepsilon.$$

Answer 2

The function $f: I\subseteq \mathbb{R}\to \mathbb{R}$ defined by $f(x)=x^{3}$ is continuous function at $x=a$ iff $$\forall \varepsilon>0\exists \delta>0\forall x\in I: |x-a|<\delta \implies |f(x)-f(a)|<\varepsilon.$$ Since, \begin{align*} |f(x)-f(a)|&=|x^{3}-a^{3}|\\ &= |x-a||x^{2}+ax+a^{2}|\\ &\leqslant |x-a|(|x|^{2}+|ax|+|a|^{2}) \end{align*} Taking $\delta\leqslant 1$ so we have $$|x-a|<1 \implies |x|<1+|a|,\quad (*)$$ Hence, \begin{align*} |x-a|(|x|^{2}+|ax|+|a|^{2})&\leqslant |x-a|\left((1+|a|)+|a|(1+|a|)+|a|^{2} \right),\quad \text{by (*)}\\ &\leqslant |x-a|(3|a|^{2}+3|a|+1) \end{align*} Then $$|f(x)-f(a)|<\varepsilon \implies |x-a|(3|a|^{2}+3|a|+1)<\varepsilon \implies |x-a|<\frac{\varepsilon}{3|a|^{2}+3|a|+1}$$ Therefore, $\delta:=\min\left\{1, \frac{\varepsilon}{3|a|^{2}+3|a|+1}\right\}$ works.

With $f: I\subseteq \mathbb{R}\to \mathbb{R}$ defined $f(x)=x^{3}$, we have that $f$ is continuous at $x=-2$. Indeed, let $\varepsilon>0$ and $\delta:=\min\left\{1,\frac{\varepsilon}{19}\right\}$, we have $$|x-(-2)|<\delta \implies |x^{3}-(-2)^{3}|=|x^{3}-(-8)|<\delta\cdot 19=\frac{\varepsilon}{19}\times 19=\varepsilon$$ so done.