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.

 How clear is this post?