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.

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.

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

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.

(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.

(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.