Let ϵ>0 be given. We wish to find δ>0 such that when ∣x−1∣<δ, ∣f(x)−1∣<ϵ.
Consider ∣f(x)−1∣<ϵ, keeping in mind we want to make a statement about ∣x−1∣:
∣f(x)−1∣<ϵ∣1/x−1∣<ϵ∣(1−x)/x∣<ϵ∣x−1∣/∣x∣<ϵ∣x−1∣<ϵ⋅∣x∣
Since x is near 1, we can safely assume that, for instance, 1/2<x<3/2. Thus
ϵ/2<ϵ⋅x.
Let δ=2ϵ. Then:
∣x−1∣<δ∣x−1∣<2ϵ∣x−1∣<ϵ⋅x∣x−1∣/x<ϵ
Assuming x is near 1, x is positive and we can bring it into the absolute value signs on the left.
∣(x−1)/x∣<ϵ∣1−1/x∣<ϵ∣(1/x)−1∣<ϵ,
which is what we wanted to prove.