Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest such δ.) What limit does this prove?

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and , we want to find a such that

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that . Then

.

Then, if , then . So, in this case, if we get that . The maximum value of delta is .

By definition, this procedure proves that . Note that f(5)=14, so this proves that f is continous at x=5.