As part of my degree, I need to be able to prove using the epsilon-delta definition of continuity that a function is continuous at a given point $c$.

This is one of those mathematical topics I will probably occasionally need to recite to myself, so, this post is to document not the technique but the reason why the technique works.

Firstly, the approach we take is to show that $f(x)-f(c) = (x-c)g(x)$. We then look for an upper bound on $g$ such that $|g(x)| \leq M$. Since the domain of $f$ could be $\mathbb{R}$, we need to narrow down the scope a little bit so we say that actually, we are looking for $|g(x)| \leq M$ for $x \in (c-a, c+a)$ given some $a > 0$.

So, we now have
$$
|f(x)-f(c)| = |(x-c)g(x)| \leq M|x-c|
$$
and we wish to be able to say that if
$$|x-c| < \delta$$
then
$$
|f(x)-f(c)| < \epsilon
$$

This now simply becomes an inequality solution problem. If $|f(x)-f(c)| < \epsilon$ then to ensure this is satisfied, we need

$$
|f(x)-f(c)| \leq M|x-c| < \epsilon
$$
Clearly, this holds when
$$|x-c| < \frac{\epsilon}{M}$$
We can use this to make a statement about the value of $\delta$ we can choose. This should satisfy:
$$|x-c| < \delta \leq \frac{\epsilon}{M}$$

If we choose delta this way, then

$$\delta > |x-c| \Rightarrow \frac{\epsilon}{M} \geq \delta > |x-c|$$
$$\frac{\epsilon}{M} > |x-c| \Rightarrow \epsilon > M|x-c|$$
$$\epsilon > M|x-c| \Rightarrow \epsilon > |f(x)-f(c)|$$

Which is a slightly laboured way to say ""it all holds together nicely""!

Now, most calculus texts also indicate we should place the restriction

$$\delta = \min\left(1, \frac{\epsilon}{M}\right)$$

on $\delta$. Why? This restricts $0 < \delta \leq 1$ which ensures we are handling a ""reasonable"" range for $|x-c|$. $1$ is a relatively safe choice, but things get ugly if we have chosen $a < 1$ since *technically* the upper bound on $g(x)$ isn't quite right for the whole range any longer.

Ok, that's what I needed to be able to explain to myself. I shall read this back in the morning to make sure it all makes sense.