Proof that $f(x) = x^2$ is continuous ($\delta-\epsilon$)? I wanted to know if there's anything wrong with my proof below.
$f(x)=x^2$ and let $c$ be an arbitrary constant.
$\mathbf1.$ 
First, $0<|x-c|<\delta$
Now,
$|x^2-c^2|=|x+c||x-c|\le(|x|+|c|)|x-c|<\epsilon\Rightarrow|x-c|<{\epsilon\over|x|+|c|}\le{\epsilon\over|x|}\Rightarrow|x-c|<{\epsilon\over|x|}$
So we have $\delta={\epsilon\over|x|}$ that satisfies the necessary condition: if $0<|x-c|<\delta$ then $|f(x)-f(c)|=|x^2-c^2|<\epsilon$
$c$ was arbitrary, so we can see that $f(x)$ is continuous, except not necessarily at $x=0$.
We need to make sure that $f(x)=x^2$ is continuous at $x=0$, since in the previous section, we can't have $|x|=0$.
$\mathbf2.$ For $x=0$, 
$0<|x-0|<\delta\Rightarrow|x|<\delta$ , and
$|x^2-0|=|x^2|=(|x|)^2<\epsilon\Rightarrow|x|<\sqrt\epsilon$
So we have $\delta=\sqrt\epsilon$ , and thus we can see that $f(x)=x^2$ is continuous at $x=0$ as well.
From conclusions drawn at the end of $\mathbf1$ and $\mathbf2$, we have shown that $f(x)$ is continuous on $x\in\mathbb R$
I just started learning about $\epsilon-\delta$. If you spot any error(s), please let me know how to correct them. Thanks!
 A: You have probably seen an $\epsilon$-$\delta$ proof before, so writing a basic proof won't tell you anything new. So I will write how I think when I prove something like this. First off, the definition:

$f(x) = x^2$ is continuous at $x_0$ if, for any $\epsilon>0$, we can find a $\delta>0$ such that for any $c$ with $|x_0-c|<\delta$, we have $|f(x_0)-f(c)|<\epsilon$. $f$ is continuous if it is continuous at every real number.

In terms of how we write the proof, that means we're given arbitrary $x_0, \epsilon$, and we must find a $\delta$ that works. You have already deduced that $|f(x_0)-f(c)| = |x_0^2 - c^2| = |x_0-c|\cdot |x_0+c|$, which is good. We will need that. Why will we need that? Because the first factor of that factorisation is the one thing we have real control over: it will be smaller than $\delta$, and we are free to choose $\delta$. We want the entire expression to be smaller than whatever $\epsilon$ we're given, so the only thing left is to make sure that the second factor, $|x_0+c|$ doesn't ruin things too much.
The standard thing to do to gain control over that term is to declare that whatever happens, I will never choose a $\delta$ which is larger than $1$ (nothing special about $1$ here, I might as well choose $1000$ or $\pi$, but I like $1$). That forces $c$ to be rather close to $x_0$, which means explicitly that $|x_0+c|<2|x_0| + 1$.
Thus, as long as we vow to never pick $\delta$ larger than $1$, that gives us
$$
|f(x_0)-f(c)| = |x_0^2 - c^2| = |x_0-c|\cdot |x_0+c|<\delta(2|x_0| + 1)\tag{*}
$$
Note that we haven't really picked a $\delta$ yet, we just know that whichever one we pick, as long as it's smaller than or equal to $1$, the inequality $\text{(*)}$ holds, which is good. Why is that good? Because ultimately, we want to be able to choose a $\delta$ so that the left-hand side is smaller than the $\epsilon$ we're given. This is easily achieved by picking a $\delta$ that makes the right-hand side of $\text{(*)}$ no bigger than $\epsilon$.
Thus we have
$$
\begin{align}
\epsilon &\geq \delta(2|x_0| + 1)\\
\delta&\leq\frac{\epsilon}{2|x_0| + 1}\tag{**}\end{align}
$$
So, which $\delta$ do we actually pick? Well, we vowed never to choose a $\delta$ greater than $1$. While under that vow, we found that any $\delta$ which fulfills $\text{(**)}$ works. Thus, if we pick exactly $\frac{\epsilon}{2|x_0| + 1}$, as long as that's smaller than $1$, and we pick $1$ if $\frac{\epsilon}{2|x_0| + 1}>1$, then we really have fulfilled both requirements, and we are done.
Explicitly, this means that we have picked $\delta = \min\left(1, \dfrac{\epsilon}{2|x_0| + 1}\right)$.
A: First error: $0<|x-c|<\delta$. You wrote that $c$ is “an arbitrary constant”. But what is $x$? And what is $\delta$?
Second error: you write that $(|x|+|c|)|x-c|<\epsilon$. What is $\epsilon$?
A: Here a correct redaction (I know it's not your question, but it's to give you an example) :
You need to prove that $$\forall a\in \mathbb R,\forall \varepsilon>0, \exists \delta>0: \forall x\in \mathbb R, |x-a|<\delta\implies |x^2-a^2|<\varepsilon.$$
So let $a\in \mathbb R$ and let $\varepsilon>0$. You have that $$|x^2-a^2|=|x-a||x+a|.$$
Since we are interested in neighborhood of $a$, we can consider $\delta<1$, (i.e. $x\in [a-1,a+1]$). So when $x\in [a-1,a+1]$, you have that $|x+a|\leq \max\{|2a-1|,|2a+1|\}$. Let denote $M=\max\{|2a-1|,|2a+1|\}$. If $\varepsilon\geq M$, set $\delta=1$ and you'll have $$|x-a|<\delta\implies |x^2-a^2|=|x-a|\varepsilon\leq \varepsilon $$
and if $\varepsilon<M$, then set $\delta=\frac{\varepsilon}{M}$, and you'll have
$$|x^2-a^2|=|x-a||x+a|\leq M|x-a|\leq M\frac{\varepsilon}{M}=\varepsilon.$$
In other word, if $\varepsilon>0$ is unspecified, set $\delta=\min\{1,\frac{\varepsilon}{M}\}$, and you'll have that $$|x-a|<\delta\implies |x^2-a^2|\leq \varepsilon,$$
what prove the claim.
A: Biggest error in my opinion:
$$\delta={\epsilon\over|x|}$$
You may not take $\delta$ to depend on $x$, but only on $c$ (the value that $x$ tends to) and on $\epsilon$.
