What is the most used method for proving continuity for simple functions such as $f(x) = x^{1/3}$ In analysis we talked about a very general definition of continuity:
$f:A \to B$ is continuous if $U \subset B$ is open, $f^{-1}(U) = V \subset A $ is open
Quite elegant
Another definition is if $(x_n) \to x \in A$, then $f(x_n) \to f(x) \in B$
Ok but you have to construct a sequence and then assume $x$ exists
Then there is the $(\epsilon-\delta) $definition, which is difficult to remember and honestly I have no clue how to use it properly

I just want to show that $f(x) = x^{1/3}$ or $f(x) = x^2$ is continuous. 
What is the most common way to prove continuity of these simple simple functions?
 A: Just for kicks, here is a completely different way to prove that $f(x)=x^2$ is continuous.  The identity function is continuous (easy to see, think for example inverse images of open sets).  Baby Rudin 4.4 teaches us that if $\lim_{x\to p} f(x)=A$ and $\lim_{x\to p} g(x)=B$, then $\lim_{x\to p} f(x)g(x)=AB$.  From this we can infer that if $f$ is continuous at $p$ (meaning $A=f(p)$) and $g$ is continuous at $p$ (meaning $B=g(p)$), then $fg$ is continuous at $p$.  Let $f$ and $g$ both be the identity function and conclude that $x^2$ is continuous.
For $x^{1/3}$ we can do the following:  $x\to x^3$ is continuous for the same reason that $x\to x^2$ is.  Baby Rudin 4.17 tells us that the inverse function of a continuous function is continuous provided the domain is compact.  Apply Baby Rudin 4.17 to $x^3$ on the compact sets $[-n,n]$ for all integers $n$, and conclude that $x^{1/3}$, the inverse function of $x^3$, is continuous at all points in $\mathbb{R}$.  
A: Let's look at $f(x) = x^2$ to get the feel for how the proofs work. Fix $\epsilon >0$ and $x \in (-L,L)$, then let's consider $y$ s.t. $| x-y | < \delta $ (you may like to think of delta as a function of $\epsilon$ and $x$ that we have to find). Then we see
$$|f(x) - f(y)| = |x^2 - y^2|=  | x-y| | x+y| \leq 2L \delta $$
Thus if we take delta like
$\delta = \frac{ \epsilon}{2L} $, we see that 
$$ |f(x) - f(y)| < \epsilon \quad \text{whenever} \quad | x-y | < \delta$$
This applies for all $\epsilon>0$ since we found $\delta(\epsilon)$.$f(x) = x^{1/3}$ is dealt with in the same manner, but instead of having a difference of squares, we have
$$  x^3 - y^3  = (x-y ) ( x^2 +xy + y^2 )$$
The trick is to get our delta bound somehow in the difference of $f(x) - f(y)$.
A: Let's review the $\delta-\epsilon$ proof.  We have
$$\begin{align}
\left|x^{1/3}-y^{1/3}\right|&=\left|(x^{1/3}-y^{1/3})\left(\frac{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}\right)\right|\\\\
&=\left|\frac{x-y}{x^{2/3}+x^{1/3}y^{1/3}+y^{2/3}}\right| \tag 1 
\end{align}$$

CASE $1$:
If $y=0$, then given $\epsilon>0$, $|x^{1/3}|<\epsilon$ whenever $|x|<\delta=\epsilon^3$.  

CASE $2$:
If $y>0$ (we leave the case for which $y<0$ to the reader), then we first choose $\delta<y/2$.  Then, $|x-y|<\delta =y/2\implies y/2 < x<3y/2$.  
Thus, we have from $(1)$ that given $\epsilon>0$
$$\begin{align}
\left|x^{1/3}-y^{1/3}\right|&=\frac{|x-y|}{(y/2)^{2/3}+(y/2)^{1/3}y^{1/3}+y^{2/3}} \\\\
&<\frac{|x-y|}{y^{2/3}}\\\\
&<\epsilon
\end{align}$$
whenever $|x-y|<\min\left(y/2,y^{2/3}\epsilon \right)$.
And we are done!
A: Here is a very special argument that applies to the function $x^\lambda$ for any $\lambda\ne0$.
Theorem. Let $U$ and $V$ be open intervals in $\Bbb R$, and $f:U\to V$ with the property that $f$ is surjective and strictly monotone, i.e. for $a,b\in U$ with $a<b$ then $f(a)<f(b)$ for an increasing function ($f(a)>f(b)$ for a decreasing function). Then $f$ is continuous.
I’ll give the proof for an increasing function, and show continuity at $x_0\in U$. Call $y_0=f(x_0)$, and let $\varepsilon>0$ be given—we may assume without loss of generality that $[y_0-\varepsilon,y_0+\varepsilon]\subset V$. Because $f$ maps $U$ onto $V$, there are elements $a,b\in U$ with $f(a)=y_0-\varepsilon$ and $f(b)=y_0+\varepsilon$. Now let $\delta=\min(x_0-a,b-x_0)$, so that $\langle x_0-\delta,x_0+\delta\rangle\subset\langle a,b\rangle$. Certainly now, if $|x-x_0|<\delta$, we get $a<x<b$, so that $f(a)<f(x)<f(b)$, i.e. $y_0-\varepsilon<f(x)<y_0+\varepsilon$, which says that $|f(x)-f(x_0)|<\varepsilon$.
(In case you prefer a proof that works by showing that the inverse image of an open is open, a quick Lemma that, under our hypotheses, the inverse image of an open subinterval of $V$ is an open subinterval of $U$ will give you the essential ingredient.)
A: I'll proceed in a standard way to show that $x^2$ is continuous.
Let $f:\mathbb{R} \to \mathbb{R}$ be defined by $f(x)=x^2$.
Let $\epsilon>0$. Then we want to show that $|f(x)-f(c)|<\epsilon$ if $|x-c|<\delta$. So we want (working "backwards"): $|x^2-c^2|=|x+c|\cdot|x-c|<\epsilon$. The problem here, is that $\delta$ cannot depend on $x$. So to remedy this problem, WLOG assume that $\delta<1$, and also that $c>0$. Then
 $$|x-c|<1 \implies -1<x-c<1 \implies c-1<x<c+1 \implies 2c-1<x+c<2c+1$$.
Then $|x+c|<2c+1$. So then, we want: $$|x+c|\cdot |x-c|<(2c+1)\cdot \delta<\epsilon$$. So we want to set $\delta=\frac{\epsilon}{2c+1}$.
Now we are ready for the "official proof":
Let $\epsilon>0$. Let $\delta=\frac{\epsilon}{2c+1}$. Suppose that $x \in \mathbb{R}$ and that $|x-c|<\delta$. Then $|f(x)-f(c)|=|x+c|\cdot|x-c|<\delta\cdot(2c+1)=\frac{\epsilon}{2c+1}\cdot(2c+1)=\epsilon$, as desired.
To aid intuition, this is almost the same thing as the "open set" definition you give:
Here: let $V$ be a neighborhood of $f(c)$. Then there exists some open ball of $f(c)$ such that $c \in B_{\epsilon}(f(c),\epsilon) \subseteq V$. Then we want there to exist some open ball $B_{\delta}(c,\delta)$ such that $f(B_{\delta}(c,\delta)) \subseteq B_{\epsilon}(f(c),\epsilon)$. But an open ball is literally defined in $\mathbb{R}$ with the standard topology to be all $y \in \mathbb{R}$ such that $|f(c)-y|<\epsilon$. All we want to show is that for each $\epsilon-ball$ there is some corresponding $\delta-ball$ such that $|x-c| \implies |f(x)-f(c)|<\epsilon$ ( or more simply that $x \in B_{\delta}(c,\delta) \implies f(x) \in B_{\epsilon}(f(c),\epsilon) \subseteq V$) , since this implies that the image of the $\delta-ball$ is in the neighborhood $V$, implying local continuity for an arbitrary point $c$.
