Which of the following statements are true?

Question

𝑓(𝑥) = { 𝑥^2 − 1, 𝑥 ≠ 1

4, 𝑥 = 1

which of the following statements are/is true?

a) the limit as x -> 1 exists

b) f(1) exists

c) f is continuous at x = 1

I said that a) and b) were correct because we know that the limit exists because we can take the derivative of x^2 -1 and we also know that at x = 1; f(x) = 4 so therefore it exists.

I was wondering if my answer was correct.

Answer 1

a) The limit does exist. For a limit to exist, you need to get the same value when you approach the input from the left and approach the input from the right.

lim x-->1- = (1)^2 - 1 = 0

lim x-->1+ = (1)^2 - 1 = 0

These are the same, so the limit does exist.

b) f(1) exists, it is defined in the function as 4.

c) f(x) is not continuous at x = 1. For the function to be continuous at a value, you need to get the same value when you approach the input from the left, approach the input from the right, AND when you plug the input into the function.

lim x-->1- = (1)^2 - 1 = 0

lim x-->1+ = (1)^2 - 1 = 0

f(1) = 4

These are not all the same, so the function is not continuous.

In response to your derivative question, it is not necessarily a good strategy to see if a limit exists by showing you can take its derivative.

The criteria for a limit existing is that you approach the same value from both sides of an input. Pretty easy to do.

The criteria for being continuous is that the limit exists AND the function equals that value when you plug in the input. This is slightly more difficult. The example above shows that the limit can exist despite the function not being continuous.

The criteria for being differentiable is that you are continuous AND you are smooth curve. This is even more difficult. The absolute value function is not differentiable at 0, since it has a sharp corner at that point. You can be continuous, but not differentiable.

Showing the function is differentiable is a clear-cut way of showing that the limit exists. But, there are plenty of functions that are not differentiable at a point, but nonetheless have a limit that exists at that point.

Answer 2

You are correct.

Since lim_x→1f(x) = lim_x→1 (x^2 - 1) = 0, the limit does exist (the fact that you can compute the derivative there is more than you need). However, since lim_x→1f(x) ≠ f(1), f is not continuous in that point. Also, f(1) clearly exists by definition