If we demand that , i.e. , then .
If we also demand that , then we get
If we let be the smallest of the two numbers and , then we have when
To be precise it has to be proven that for any real number , a real number can be found such that the implication holds.
So the epsilon-delta game begins with epsilon.
Now, for arbitrarily chosen look at the inequality . Treat the expression as you would any other normal variable (think of it as ) and you have a quadratic inequality which holds when .
To put it all in one place you have:
For arbitrarily selected , you can use any and the inequality would imply that the inequality holds. Thus you have proven the continuity at 3.
What Mondreus used as an example is simply a special case of this story: were you to choose then you should use