Definition:

A function f(x) is continuous at a given point x = a if those three conditions below are met “simultaneously”:

(i) f(a) is defined. (i.e; a is in the domain of f)

(ii) \lim\limits_{x \to a} f(x) exists.

(iii) \lim\limits_{x \to a} f(x) = f(a).

NOTE:

  • If any one of the three conditions is false, then f is discontinuous at a, or it has a discontinuity at a.

Let’s now look at the different cases where f(x) may not be continuous at x = a.

(i) f(a) is defined but \lim\limits_{x \to a} f(x) does not exist.

JumpDiscontinuity

At a = 0, the function is not continuous despite f(a) is defined (Here, f(a) is equal to -1). This is because the two one-sided limits are not equal and as a consequence, the limit does not exist. This is called a jump discontinuity.

(ii)  \lim\limits_{x \to a} f(x) exists, but f(a) is not defined.

RemovableDiscontinuity

Assume a is the x-value where there is a hole in the graph. We can see that the limit from the right of a and the limit from the left of a are equal. Therefore, the limit exists. However, f is undefined at a. This is another form of discontinuity, and it known as a removable discontinuity.

(iii) f(a) is defined, \lim\limits_{x \to a} f(x) exists, but \lim\limits_{x \to a} f(x) \neq f(a).

RemovableDiscontinuity1

Assume a is the x-value where there is a hole in the graph. This one is also an example of removable discontinuity. However, there is a slight difference:

Here f(a) is defined and \lim\limits_{x \to a} f(x) exists. The reason for the discontinuity is because \lim\limits_{x \to a} f(x) \neq f(a).

We have seen different examples of how a function depends on the conditions for it to be continuous. If one of the conditions fail, the function is discontinuous.

Let’s now work through some examples.

E.g; (a) Given the graph of f(x), determine if f(x) is continuous at x = -4, x = -1, x = 2 and x = 4.

Example1

(i) At x = -4, f(-4) is defined and is equal to 3. However, \lim\limits_{x \to -4^{+}} f(x) \neq \lim\limits_{x \to -4^{-}} f(x). Or, in simple terms the limit as x approaches -4 from the left is not equal to the limit as x approaches -4 from the right. Therefore, the function has a discontinuity at the point x = -4.

(ii) Now, at x = -1, we see that f(-1) is defined, the \lim\limits_{x \to -1^{+}} f(x) \equiv \lim\limits_{x \to -1^{-}} f(x) and \lim\limits_{x \to -1} f(x) \equiv f(-1). We can say that the function is continuous at x = -1.

(iii) When x = 2, f(2) is defined and is equal to -1. However, the limit as x approaches 2 from the left is not equal to the limit as x approaches 2 from the right. Clearly, one of the conditions has failed. Hence, we know that f is not continuous at x = 2.

(iv) At x = 4, despite \lim\limits_{x \to 4^{+}} f(x) \equiv \lim\limits_{x \to 4^{-}} f(x), f(4) is undefined. Therefore, the function is discontinuous at x = -4.

(b) Given the graph of f(x), determine if f(x) is continuous at x = -8, x = -2, x = 6 and x = 10.

Example2

(i) At x = -8, it is clear that f(-8) is defined and \lim\limits_{x \to -8^{+}} f(x) \equiv \lim\limits_{x \to -8^{-}} f(x). However, \lim\limits_{x \to -8} f(x) \neq f(-8) as f(-8) is equal to -3. Therefore, the graph is discontinuous at x = -8.

(ii) Now, at x = -2, we can see that f(-2) is defined, but \lim\limits_{x \to -2^{+}} f(x) \neq \lim\limits_{x \to -2^{-}} f(x). In fact, we can say that \lim\limits_{x \to -2^{-}} f(x) \equiv f(-2) and \lim\limits_{x \to -2^{+}} f(x) = \infty. As the right-sided limit and the left-sided limit are not equal, the graph has a discontinuity at x = -2.

(iii) When x = 6, \lim\limits_{x \to 6^{+}} f(x) \neq \lim\limits_{x \to 6^{-}} f(x). Even though f(6) is defined, the function is not continuous here either, at x = 6.

(iv) At x = 10, f(10) is defined, \lim\limits_{x \to 10^{+}} f(x) \equiv \lim\limits_{x \to 10^{-}} f(x) and \lim\limits_{x \to 10} f(x) = f(10). Therefore, we can conclude that it is continuous at this point.

How clear is this post?