Consider the function f defined by f(x)=8x−3. Which quantities
would need to be related to show that limx→1f(x)=5 using the
formal definition of the limit?
We must show that given any ε > 0 there exists δ such that
whenever |x-1| < δ, then |(8x-3)-5| < ε.
Proof:
Suppose we are given ε > 0
|(8x-3)-5| < ε if and only if
|8x-3-5| < ε if and only if
|8x-8| < ε if and only if
|8(x-1)| < ε if and only if
8|x-1| < ε if and only if
|x-1| < ε/8
So we take δ = ε/8.
We can stop here, but you can continue and say:
Now we have proved that if we take any ε > 0, then there exists
δ = ε/8 such that whenever
|x-1| < ε/8, then
8|x-1| < ε, then
|8(x-1)| < ε, then
|8x-8| < ε, then
|8x-3-5| < ε, then
|(8x-3)-5| < ε
But usually, you don't need to go through that second part if
you will put "if and only if" between the steps of the first part
as I did above.
Edwin
