Using the definition of limit show that $\lim_{x→x_0} x^3=x_0^3$

Question

We want to show that $∀ϵ>0,∃δ_ϵ(x_0)>0$ such that if $0<|x−x_0|<δ_ϵ(x_0)$, then $|x^3−x_0^3|<ϵ$.

$|x^3−x_0^3|=|(x-x_0)(x+x_0)^2|=|x-x_0||x^2 + xx_0 + x_0^2|$. Here we have $|x-x_0|<δ_ϵ(x_0)$.

Assume $δ_ϵ(x_0)=1$. Then, $|x-x_0| < 1$, and $|x|-x_0 ≤ |x-x_0| < 1$, implying that $|x| < x_0 +1$, implying that $$|(x^2 + xx_0 + x_0^2)| ≤ |x|^2 + |x|x_0 + x_0^2 < 3x_0^2 + 3x_0 + 1$$.

So, take any ϵ>0, then $$|x^3−x_0^3|<ϵ$$ if,

$$|x-x_0| < \frac{ϵ}{3x_0^2 + 3x_0 + 1}$$.

Let $$δ_ϵ(x_0) = min\{1, \frac{ϵ}{3x_0^2 + 3x_0 + 1}\}$$ and we are done.

Could you please verify my proof?

Answer 1

Hint: It is $$x^3-x_0^3=(x-x_0)(x^2+xx_0+x_0^2)$$