Prove that the given function: $h(x) = x^2$ is continuous at every real number Use the definition of continuity to prove that the given function: $h(x) = x^2$ is continuous at every real numbers.
So far for my proof I have:
Let $\epsilon >0$ be given to us. We must show there exists a $\delta >0$ such that $$|x-c| < \delta \Rightarrow |x^2-c^2| < \epsilon.$$
This is where I run into a problem trying to figure out what to choose for $\delta$ to be equal too.
For my scratch work I have:
$|x^2-c^2| = |(x-c)(x+c)| = |x-c||x+c|$  (I know that the part giving me the problem is $|x+c|$)
Let $|x-c| <1 $
Then,
$|x+c| = |x-c+2c|\leq |x-c|+2|c| < 1+ |2c|$
This is where I get stuck I do not know where to go from here
 A: You had found that $$|x^2 -c^2| < |x-c| (1 + 2|c|)$$ for any $c$ and $x$ such that $|x-c| <1$.
Given $\epsilon >0$, define $\delta :=\min\{1,\frac{\epsilon}{1+2|c|}\}$ (Note that denominator in second term is always positive.) We have :
\begin{align} |x-c| < \delta \implies & |x-c| < 1  \, \text{and}   \\ & |x-c| < \frac{\epsilon}{1+2|c|} \end{align}
This means that \begin{align}|x^2 -c^2| &< |x-c| (1 + 2|c|)\\ &< \frac{\epsilon}{1+2|c|} (1 + 2|c|) = \epsilon  \end{align}
A: If $x- c\approx 0$ then $x \approx c$ and $|x + c| \approx 2|c|$ and as $c$ is constant for evaluating AT $x = c$ we are allowed to (although I prefer not to) to use $c$ if we are attempting to prove $f$ is continuous at $x=c$.  (The thing is, and this is the reason I prefer not to, is that we need a different $\delta$ in terms of $c$ for every point we try to evaluate the continuity of $x$ at.  But that is acceptable).
SO to put the argument that $x-c \approx 0$ etc.  in valid terms of $\delta$....
Claim:  If $|x-c| < \delta$ then $|x+c| < 2|c| + \delta$.
Argument:
If $|x-c| < \delta$ then $-\delta < x- c < \delta$ and $c-\delta < x  < c+\delta$.  SO $2c - \delta < x+c < 2c +\delta$.
If $c \ge 0$ then $-2c - \delta < 2c - \delta < x+c < 2c + \delta$ and $|x+c| < 2|c| +\delta$.
If $c < 0$ then $-2|c| - \delta < x+ c < -2|c| + \delta < 2|c| + \delta$ and $|x+c| < 2|c| + \delta$.
So either way: $|x+c| < 2|c| + \delta$.
Good enough.
.......
So if $|x-c| < \delta$ then $|(x-c)(x+c)| < \delta(2|c| + \delta) = \delta^2 + 2|c|\delta$.
If we assume $\delta \le 1$ then $\delta^2 \le \delta$ and $\delta^2 + 2|c| \delta \le  (1+2|c|)\delta$.
So we want for any $\epsilon > 0$ then $(1+2|c|)\delta \le \epsilon$ so let $\delta = \min (1, \frac {\epsilon}{1+ 2|c|})$ and we are golden.
If $|x-c| < \delta$ then $|x^2 -c^2| = |x-c||x+c|< \delta (\delta + 2|c|)\le \delta^2 + 2|c|\delta \le (1+2|c|)\delta \le \epsilon$.
