Definitions of continuity seem to contradict.

Question

A function f(x) is considered continuous at c if: the limit of f(x) from the left (of c) equals the limit from the right, which equals f(c). But another definition is that if the derivative exists at f(c), the function is continuous at c. However, consider the function (1-x)/(1-x). Clearly this function is not continuous at x = 1 because you get 0/0. But we can take the derivative of this function, which exists…and equals 0. So is it continuous or not at x = 1? Seems like these two definitions of continuity contradict?

Answer 1

I'm expanding on what Kenneth said to help explain.

The derivative of f(x) = (1-x)/(1-x) does not exist at x = 1.

Using the Quotient Rule, we calculate that:

• f'(x) = [-(1-x) + (1-x)]/(1-x)² = 0/(1-x)²

We gotta be careful whenever we're eliminating variable stuff from denominators.

0/(1-x)² ≠ 0 for all x,

because to say it did would be assuming that (1-x)² ≠ 0, which is not always true, right?

So at x = 1, we get that (1-x)² = 0, so the derivative solution of 0 does not apply there.

Plugging in x = 1, we get

• f'(1) = 0/(1-1)² = 0/0, or DNE

Yes, the limit exists at x = 1, but not the derivative. So f(x) is not continuous at x = 1. Very good question Mark!

Answer 2

fx) = (1-x)/(1-x) has the simplified value x = 1 over its domain, which excludes x=1. It has no derivative at x=1 because of its definition (includes domain restriction). Nor is it continuous at x=1, because f(1) DNE. 

 

There are no contradictions.