prove $x \mapsto x^2$ is continuous I am to show the continuity of this function with the help of $\epsilon$-$\delta$ argument. 
The function is: $g: \Bbb{R} \rightarrow \Bbb{R}$, $x \mapsto x^2$. 
Given the $\epsilon$-$\delta$ definition of limit, I tried as follows: 
We must have: $|f(x)-f(x_0)|<\epsilon$, so then $|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|$, so $|x-x_0|<\frac{\epsilon}{|x+x_0|}$. So now I will choose $\delta=\frac{\epsilon}{|x+x_0|}$. 
Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively.
 A: No. $\delta$ must only depend on $x_0,\epsilon$ and never on $x$. Here is what we will do:
The problematic term is $\left|x+x_0\right|$. We have that 
\begin{equation} \left|x+x_0\right|= \left|x-x_0+2x_0\right|\le \left|x-x_0\right|+\left|2x_0\right|<\delta+2\left|x_0\right|\end{equation}
Thus, 
\begin{equation}\left|x-x_0\right|<\delta\implies \left|f(x)-f(x_0)\right|=\left|x+x_0\right|\left|x-x_0\right|<(\delta+2\left|x_0\right|)\delta\end{equation}
We must choose a $\delta$ so that
$$(\delta+2\left|x_0\right|)\delta<\epsilon$$
Choosing a $\delta$ by the above might be complicating. But because $\delta$ is ours to choose we can simplify things a bit by demanding $\delta<1$. Then,
$$(\delta+2\left|x_0\right|)\delta<(1+2\left|x_0\right|)\delta$$
and so it suffices to choose $0<\delta<1$ so that
$$(1+2\left|x_0\right|)\delta<\epsilon$$
Things should be straighforward now.
For the sake of completion we have 
$$(1+2\left|x_0\right|)\delta<\epsilon\iff \delta<\frac{\epsilon}{1+2\left|x_0\right|}$$
But because $\delta<1$ we must choose $\delta>0$ so that
$$\delta<\min\left\{1,\frac{\epsilon}{1+2\left|x_0\right|}\right\}$$
Taking 
$$\delta=\frac12\min\left\{1,\frac{\epsilon}{1+2\left|x_0\right|}\right\}$$
completes the proof
A: Assume we found
$\delta$,
$|x|\leq |x-x_0+x_0|\leq|x-x_0|+|x_0|<\delta+|x_0|.$
If this $\delta$ works, then any $\delta'\leq \delta$ also works, therefore we may assume $\delta<1$ as well.
Now we have an estimate of the form $|x|<1+|x_0|$. One can see that we just restricted $x$ in the open interval $(x_0-1,x_0+1)$.
Now, we have
$$|f(x)-f(x_0)|=|x^2-x_0^2|\leq (1+2|x_0|)|x-x_0|$$.
One can observe that $\delta<\frac{\epsilon}{(1+2|x_0|)}.$
Since we also want to estimate first, namely $|x|<1+|\delta|$.
Therefore we choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$
Coming to the formal proof;
Fix $x_0\in\Bbb{R}$, given  $\epsilon>0$. Choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$
Let $x\in\Bbb{R}$ be such that $|x-x_0|<\delta$. Then $|x-x_0|<1, |x|=|x-x_0+x_0|\leq |x-x_0|+|x_0|<1+|x_0|$.
Now,$|f(x)-f(x_0)|=|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|x-x_0||x+x_0|\leq |x-x_0|(1+2|x_0|)<(1+2|x_0|)\delta<\epsilon.$
