Proof that $f(x) = x^3$ is continuous?

Question

I recently asked about proof that $x^2$ is continuous, and after the feedback I received, I decided to see if I learned correctly using another example: $f(x)=x^3$. I'm going to have some scratch work before the actual proof. Below, $c$ is an arbitrary constant and I want to prove that $f(x)=x^3$ is continuous at $x=c$ to show that $x^3$ is continuous everywhere. $x_0$ is an arbitrary point in $(c-\delta,c+\delta)$.

Scratch work (lots of algebra):

$0<|x_0-c|<\delta$

$|f(x_0)-f(c)|=\left|x_0^3-c^3\right|=|x_0-c|\left|x_0^2+cx_0+c^2\right|\le|x_0-c|\left(|x_0|^2+|cx_0|+\left|c^2\right|\right)$

side note:

If we let $\delta\le1$, then $|x_0-c|<1\Rightarrow|x_0|=|x_0-c+c|\le|x_0-c|+|c|<1+|c|$

$\Rightarrow|x_0|<1+|c|$

\end of side note

Picking up from the place before the side note, we have

$|x_0-c|\left(|x_0|^2+|cx_0|+\left|c^2\right|\right)<|x_0-c|\left(\left(1+|c|\right)^2+|c|(1+|c|)+\left|c^2\right|\right)$

$= |x_0-c|\left(1+2|c|+|c|^2+|c|+\left|c^2\right|+\left|c^2\right|\right)=|x_0-c|\left(3\left|c^2\right|+3|c|+1\right)$

So far we have $|x_0-c|\left(3\left|c^2\right|+3|c|+1\right)<\epsilon\Rightarrow|x_0-c|<\frac{\epsilon}{3\left|c^2\right|+3|c|+1}$

And we can have $\delta={\epsilon\over3\left|c^2\right|+3|c|+1}$

/end of scratch work

Proof:

If $0<|x_0-c|<\delta$ and $\delta\le1$ and $\delta={\epsilon\over3|c^2|+3|c|+1}$, then

$<|x_0-c|\left(3\left|c^2\right|+3|c|+1\right)$

$<\delta*(3\left|c^2\right|+3|c|+1)=\frac{\epsilon}{3|c^2|+3|c|+1}\left(3\left|c^2\right|+3|c|+1\right)=\epsilon$

$\Rightarrow\left|x_0^3-c^3\right|<\epsilon$

Thus if $\delta=\min\left\{1,{\frac{\epsilon}{3\left|c^2\right|+3|c|+1}}\right\}$, then $\left|x_0^3-c^3\right|<\epsilon$, so $x^3$ is continuous, and we are done.

I appreciate your criticisms. If you spot any error(s), please let me know. Thanks!

Not a criticism, but a suggestion to make things easier. Just prove that every polynomial $f(x)$ is a continuous function - see this MSE-question. — Dietrich Burde, Aug 01 '17 at 12:08
Can't you just use "the product of continuous functions is continuous"? $g(x) = x$ is continuous, $f(x) = g(x) \times g(x) \times g(x)$? — user8469759, Aug 01 '17 at 12:10
I'm using these as exercises because I only recently became interested in epsilon-delta problems such as these. Thanks for your suggestion; I'll look into that later as well. — orchid, Aug 01 '17 at 12:10
@user8469759 My goal isn't to prove that x^3 is continuous efficiently, but to prove it using epsilon-delta as practice. — orchid, Aug 01 '17 at 12:11
You should try proving that if $f$ and $g$ are continuous, so is $f\cdot g$ (using an $\epsilon$-$\delta$ proof) — MPW, Aug 01 '17 at 12:13
This is a very good exercise! You should also look into the epsilon-delta proof that $1/x$ is continuous =) — étale-cohomology, Aug 01 '17 at 12:16
A sketch of a proof from the definition is given at Continuity of $f(x)=x^n$. — Andrew D. Hwang, Aug 01 '17 at 14:14
Could you please explain why you choose $\delta$ to be the minimum of the two? — stoic-santiago, Aug 18 '19 at 05:43

Proof that $f(x) = x^3$ is continuous?

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