Is my proof of $\lim_{x\rightarrow c}x^2=c^2$ correct? I know the most common proof of $\lim_{x\rightarrow c}x^2=c^2$. But I wonder if my alternative proof is valid and correct. Here's my proof.
Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$
For the convenience for observation, suppose that $0<|x-c|<\square$, we want to find out which $\square$ is ok, and then know what $\delta$ to pick. Since $|x^2-c^2|<\varepsilon\Leftrightarrow |x+c||x-c|<\varepsilon$ and 
\begin{alignat*}{3}
&0<|x-c|<\square\\
\Longleftrightarrow &c-\square<x<c+\square&(except\ x=c)\\
\Longleftrightarrow &2c-\square<x+c<2c+\square\qquad&(except\ x=c)\\
\Longrightarrow &|x+c|< |2c|+\square
\end{alignat*}
So we see that if we want  $|x+c||x-c|<(|2c|+\square)\square<\varepsilon$ to be true, we need to solve a positive solution of the quadratic inequality $\square^2+|2c|\square-\varepsilon<0$, by the relationship between roots and coefficient, we know $\square^2+|2c|\square-\varepsilon=0$ has exactly one positive and one negative root. Hence the solution of the prior inequality is $(-c-\sqrt{c^2+\varepsilon},-c+\sqrt{c^2+\varepsilon})$, where the right endpoint is positive. Thus, we pick $\delta=-c+\sqrt{c^2+\varepsilon}$ to complete the proof.
 A: The general idea is great, except I'd be careful about the last $\iff$, since the converse isn't necessarily true. It's safer to use a $\Longrightarrow$, since that's all we really need. Indeed, if $c = 7$ and $\square = 1$, then:
$$
13 < x + 1 < 15 \implies -15 < x + 1 < 15 \iff |x + 1| < 15
$$
But the converse is not true, since if $x = 4$, then $|x + 1| < 15$ is true but $13 < x + 1 < 15$ is false.

Here's a cleaned up version of your proof, organized a bit differently:

Given any $\varepsilon > 0$, let $\delta = \sqrt{c^2 + \varepsilon} - |c|$, which is certainly positive. Then if $0 < |x - c| < \delta$, observe that:
  \begin{align*}
|x^2 - c^2|
&= |x - c||x + c| \\
&= |x - c||(x - c) + 2c| \\
&\leq |x - c|(|x - c| + |2c|) &\text{by the triangle inequality} \\
&< \delta(\delta + 2|c|) \\
&= (\sqrt{c^2 + \varepsilon} - |c|)(\sqrt{c^2 + \varepsilon} + |c|) \\
&= (c^2 + \varepsilon) - c^2 \\
&= \varepsilon
\end{align*}
  as desired. $~~\blacksquare$

A: You have picked $\delta>0$ such that $\delta=-c+\sqrt{c^2+\varepsilon}$
This is a problem. Let c=-5.
so $\delta=5+\sqrt{25+\varepsilon}$
substitute $\varepsilon=0.0001$.
So $\delta$ is approximately equal to 10.
This is where the problem lies. So x can range from -15 to +5. Let x=0. So $x^2$ =0. But 0 does not into (25-$\varepsilon$,25+$\varepsilon$) or (25-0.0001,25+0.0001). So, there is some x satisfying $0<|x-5|<\delta$ for which f(x) does not line in  (25-0.0001,25+0.0001).
Hence a contradiction.
So, we can't use this $\delta$.
