Proving $\lim_{x \to c} x^2 = c^2$ using $\epsilon , \delta$

Question

I am trying to construct a proof for $\lim_{x \to c} x^2 = c^2$. Here is my attempt. Let |x-c| < $\delta$ for some $\delta > 0$. $|x^2 - c^2| = |x+c||x-c|.$ We have hat |x-c| < $\delta$. Let $\delta$ = 1 so that -1 < |x-c| < 1 or -1+c < x < c+1. Since c+1 is an upper bound on x we have |x+c||x-c| $< |c+1+c||x-c| = |2c+1||x-c|.$ We can state that |2c+1||x-c| < (2c+1)*$\frac{\epsilon}{2c+1} = \epsilon$. So $\lim_{x \to c} x^2 = c^2$. Does this look right? Is my step in obtaining an upper bound on x+c correct? Thanks!

Answer 1

You are going to be in big trouble if $c = -\dfrac 12$. Instead use the triangle inequality to get $$|x-c| < 1 \implies |x + c| = |x-c + 2c| \le |x-c| + 2|c| \le 1 + 2|c|.$$

Answer 2

The problem is that $|2c+1|$ could be $0$ if $c = -1/2$, and then no division by it is legitimate.

A usual trick to bound away the annoying $|x+c|$ is to use a preliminary bound for $|x-c|$. In this case, you might use $|x-c| < 1$, which gives $|x| - |c| \leq ||x| - |c|| \leq |x-c| < 1$, so $|x+c| \leq |x| + |c| < 1 + 2|c|$. Then, given any $\varepsilon > 0$, we have $|x^{2} - c^{2}| < \varepsilon$ if $|x-c| < 1$ and $\varepsilon/(1+2|c|)$ (note that here $1+2|c|$ cannot be $0$ whatever $c$ is); hence $\delta := \min \{ 1, \varepsilon/(1+2|c|) \}$ suffices.