First, see the comment of Brian Moehring, following the original question. This indicates that the demonstration below, that $f(x) = x^{(1/3)}$ is a continuous function, is not pertinent.
So much analysis there, that I can't bring myself to delete it. However, I have added an Addendum to address the issue of the continuity of $(x^3).$
Responding to a comment/question from the OP:
Given $f:\Bbb{R}\to \Bbb{R}$ such that $f(x) = x^{(1/3)}.$
Demonstration that for all $x_0 \in \Bbb{R}, ~f(x)~$ is continuous at $x = x_0$:
Personally, I found the answers given here, here, and here to be somewhat sophisticated. It seems reasonable to (instead) provide a (primitive) $\epsilon, \delta$ proof.
Definition of continuity:
- $f(x)$ is defined at $(x = x_0)$.
- $\forall \epsilon > 0, ~\exists \delta > 0~$ such that
$0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon.$
Clearly, the first constraint above is satisfied, for all $x_0 \in \Bbb{R}.$
What the second constraint is saying, in effect, is that for each $(x_0)$, there exists a fixed finite limit $L$, such that $\lim_{x\to x_0} f(x) = L$ and that in all cases, $L = f(x_0)$.
Intermediate Results
Lemma 1:
$a < b \iff a^3 < b^3$.
Proof:
$g(x) = x^3 \implies g'(x) = 3x^2.$
Therefore, $g(x)$ is a strictly increasing function, throughout $\Bbb{R}.$
Lemma 2:
$~~~\displaystyle 0 < \delta < x_0, ~~~0 < \epsilon < x_0^{(1/3)},
~~~ \delta < \left[3x_0^{(2/3)}\epsilon - 3x_0^{(1/3)}\epsilon^2\right] ~~~\implies $
$~~~\displaystyle \left[x_0^{(1/3)} - \epsilon\right] < (x_0 - \delta)^{(1/3)}.$
$~~~\displaystyle 0 < \delta < x_0, ~~~0 < \epsilon,
~~~ \delta < \left[3x_0^{(2/3)}\epsilon\right] ~~~\implies $
$~~~\displaystyle \left[x_0^{(1/3)} + \epsilon\right] > (x_0 + \delta)^{(1/3)}.$
Proof:
Using Lemma 1, I can show (for example) that $R < S$ by showing that $R^3 < S^3.$
$\displaystyle \left[x_0^{(1/3)} - \epsilon\right]^3
= x_0 - 3x_0^{(2/3)}\epsilon + 3x_0^{(1/3)}\epsilon^2 - \epsilon^3.$
By assumption, this is less than $(x_0 - \delta).$
Similarly,
$\displaystyle \left[x_0^{(1/3)} + \epsilon\right]^3
= x_0 + 3x_0^{(2/3)}\epsilon + 3x_0^{(1/3)}\epsilon^2 + \epsilon^3.$
By assumption, this is greater than $(x_0 + \delta).$
Proof of Second Constraint
From the premise, you have that
$$- \delta < (x - x_0) < \delta \iff (x_0 - \delta) < x < (x_0 + \delta).\tag{1}$$
From the conclusion you want to establish that
$$ - \epsilon < x^{(1/3)} - x_0^{(1/3)} < \epsilon \iff
\left[x_0^{(1/3)} - \epsilon\right] < x^{(1/3)} < \left[x_0^{(1/3)} + \epsilon\right].\tag2$$
$\underline{\text{Case 1:}~x_0 > 0.}$
Use the following specifications:
$$
\text{If} ~~\left[x_0^{(1/3)} > \epsilon\right]
~~\text{then}~~
\delta = \min\left\{ ~\frac{x_0}{2}, ~\left[x_0^{(2/3)}\epsilon - x_0^{(1/3)}\epsilon^2\right] ~\right\}.$$
$$
\text{Otherwise}~~
\delta = \min\left\{ ~\frac{x_0}{2}, ~\left[x_0^{(2/3)}\epsilon\right] ~\right\}.$$
It is assumed that $\epsilon > 0$ and that $\delta$ is as specified above.
From this, you know that $x_0 - \delta > 0.$
Using the specification, Lemma 1 and inequality 1 above, you have that
$\displaystyle 0 < (x_0 - \delta)^{(1/3)} < x^{(1/3)} < (x_0 + \delta)^{(1/3)}.$
Using the specification and Lemma 2, you know immediately that
$\displaystyle \left[x_0 + \delta\right]^{(1/3)} < \left[x_0^{(1/3)} + \epsilon\right].$
If $~~x_0^{(1/3)} - \epsilon \leq 0,~~$ then this immediately establishes that
$\displaystyle x_0^{(1/3)} - \epsilon < \left[x_0 - \delta\right]^{(1/3)}.$
Otherwise, the specification, with Lemma 2, establishes that
$\displaystyle x_0^{(1/3)} - \epsilon < \left[x_0 - \delta\right]^{(1/3)}.$
Therefore, the specification for $\delta$ establishes inequality (2) above.
$\underline{\text{Case 2:}~x_0 < 0.}$
Define $~y_0 = -x_0 \implies $
- $y_0 > 0.$
- $x_0^{(1/3)} = - y_0^{(1/3)}.$
For all $~x < 0,~$ define $~y = -x \implies $
- $y > 0.$
- $x^{(1/3)} = - y^{(1/3)}.$
Assume that $\epsilon > 0$.
Further assume that Case 1 is invoked, so that $\delta$ is specified so that whenever
$$- \delta < (y - y_0) < \delta \iff (y_0 - \delta) < y < (y_0 + \delta)\tag{3}$$
then, the following is satisfied:
$$ - \epsilon < y^{(1/3)} - y_0^{(1/3)} < \epsilon \iff
\left[y_0^{(1/3)} - \epsilon\right] < y^{(1/3)} < \left[y_0^{(1/3)} + \epsilon\right].\tag4$$
Then, inequality 3 above is satisfied if and only if
$|y - y_0| < \delta \iff |x - x_0| < \delta.$
Further, inequality 4 above is satisfied is and only
$\left|y^{(1/3)} - y_0^{(1/3)}\right| < \epsilon \iff
\left|x^{(1/3)} - x_0^{(1/3)}\right| < \epsilon.$
Therefore, continuity is established for all $x_0 < 0.$
$\underline{\text{Case 3:}~x_0 = 0.}$
Set $\displaystyle \delta = \frac{1}{2}\epsilon^3 \implies \delta^{(1/3)} < \epsilon,$ by Lemma 1.
Then, using Lemma 1 repeatedly,
$-\delta < x < \delta \implies$
$- \epsilon < -\delta^{(1/3)} < x^{(1/3)} < \delta^{(1/3)} < \epsilon.$
Therefore, continuity is established for $(x_0 = 0).$
Addendum
Let $h(x) = x^3.$
Prove that $h(x)$ is continuous everywhere.
First of all, I strongly suspect that this has been proven somewhere on MathSE. However, I could not find such a proof, which indicates that the OP will also have trouble finding such a proof.
Therefore, I will provide a proof below.
Analysis will be very similar to the analysis in the original answer.
Given:
$$- \delta < (x - x_0) < \delta \iff (x_0 - \delta) < x < (x_0 + \delta).\tag{5}$$
To Prove:
$$ - \epsilon < x^3 - x_0^3 < \epsilon \iff
\left[x_0^3 - \epsilon\right] < x^3 < \left[x_0^3 + \epsilon\right].\tag6$$
This proof is much easier than the previous proof.
$\underline{\text{Case 1:} ~x_0 > 0.}$
Use the following specification:
$$
\delta = \min\left\{ ~\frac{x_0}{2}, \frac{1}{2}, ~\left[\frac{\epsilon}
{2 \times (3x_0^2 + 3x_0 + 1)}\right]~\right\}.$$
It is assumed that $\epsilon > 0$ and that $\delta$ is as specified above.
From this, you know that $x_0 - \delta > 0.$
Using the specification, Lemma 1 and inequality 5 above, you have that
$\displaystyle 0 < (x_0 - \delta)^3 < x^3 < (x_0 + \delta)^3.$
Further, since $\delta < 1$,
$(x_0 + \delta)^3 = x_0^3 + 3x_0^2\delta + 3x_0\delta^2 + \delta^3$
$< x_0^3 + \delta(3x_0^2 + 3x_0 + 1) < x_0^3 + \epsilon.$
Also, you have that
$\epsilon > \delta(3x_0^2 + 3x_0 + 1)$
$> 3x_0^2 \delta + 3x_0\delta^2 + \delta^3$
$> 3x_0^2 \delta - 3x_0\delta^2 + \delta^3$.
This implies that
$x_0^3 - \epsilon < x_0^3 - 3x_0^2 \delta + 3x_0\delta^2 - \delta^3 = (x_0 - \delta)^3.$
Therefore, when $x_0 > 0$, inequality (6) above is established, so $h(x) = x^3$ is continuous at $x_0$.
$\underline{\text{Case 2:} ~x_0 < 0.}$
Adopting the nomenclature of Case 2 in the original answer, the analysis is virtually identical to that of Case 2.
That is,
$x_0^3 = -y_0^3.$
$x^3 = -y^3.$
$|x - x_0| = |y - y_0|$.
$|x^3 - x_0^3| = |y^3 - y_0^3|.$
Therefore, $h(x) = x^3$ is also continuous at $x_0 < 0.$
$\underline{\text{Case 3:} ~x_0 = 0.}$
Use the following specification:
$$
\delta = \min\left\{ \frac{1}{2}, ~\left[\frac{\epsilon}
{2}\right]~\right\}.$$
It is assumed that $\epsilon > 0$ and that $\delta$ is as specified above.
Using the specification, Lemma 1 and inequality 5 above, you have that
$\displaystyle - \delta^3 < x^3 < \delta^3.$
Further, since $\delta < 1, ~~\delta^3 < \delta < \epsilon.$
Therefore,
$-\epsilon < -\delta < -\delta^3 < x^3 < \delta^3 < \delta < \epsilon.$
Therefore, when $x_0 = 0$, inequality (6) above is established, so $h(x) = x^3$ is continuous at $x_0 = 0.$
