Proof Of The Product Law

Lets say we have these two limits:

This means:

Before launching into the actual proof, let's do a little Algebra. First, expand the following product:

This gives us a new way of defining f(x)g(x), we will use that here:

Since "lim_x->af(x)-L = 0" and "lim_x->ag(x)-M = 0", then "lim_x->a[f(x)-L][g(x)-M] = 0" (the proof of this will be shown later). We can replace "lim_x->a[f(x)-L][g(x)-M]" with 0:

This means:

Our proof works if lim_x->a[f(x)-L][g(x)-M] = 0 is true, now lets try to proof this. For every number ε > 0, there is a number δ₁ > 0 and a number δ₂ > 0 such that:

Let δ = min{δ₁, δ₂}:

This would mean lim_x->a[f(x)-L][g(x)-M] = 0.