Prove $|x-c|<\delta \implies |x^2-c^2|<\epsilon$

Question

I would like to know if my $\epsilon-\delta$-proof of the following is correct. $c$ is given and I'm trying to show that $\forall e > 0, \exists d > 0$ so that: $$ |x-c| < d \implies |x^2-c^2| < e $$ First I expand the squares: $$ |x-c| < d \implies |x-c||x+c| < e $$ By using the triangle inequality on $|x-c|$ I bound $|x|$ $$ |x|-|c| \leq |x-c| \land |x-c|<d\implies |x|< d + |c| $$ I then use the triangle inequality again on the $|x+c|$ factor $$ |x+c| \leq |x| + |c| < d + 2|c| $$ Putting the bounds together $$ |x+c||x-c| < d(d+2|c|) = d^2 + 2d|c| $$ I then set the upper bound equal to $e$ and solve for $d$: $$ d^2 + 2d|c| = e \iff d = \sqrt{e+|c|^2} - |c| $$ Since I found a $d$ as a function of $e$ and $c$, is my proof complete?

Answer 1

Your argument looks fine. Expect that:

1. You might want to check your $d$ is really positive.

2. Your last $\Leftrightarrow$ is wrong. A quadratic equation should have two solutions.