Definition:

(i) A function $f$ is said to be continuous from the right at a if

$\lim\limits_{x \to a^{+}} f(x) = f(a)$

We can see that, as the function approaches a certain x-value from the right, $f$ is defined and

$\lim\limits_{x \to a^{+}} f(x) \equiv f(a)$

And as the function approaches a certain x-value from the left, $f$ is not defined, i.e;

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

Therefore, we say that the function is continuous from the right at this point, but is discontinuous from the left.

(ii) A function $f$ is said to be continuous from the left at a if

$\lim\limits_{x \to a^{-}} f(x) = f(a)$

Here, it is clear from the graph that the function is continuous from the left as approaches 3. This is because the function is defined at x = 3 and,

$\lim\limits_{x \to 3^{-}} f(x) \equiv f(3)$

However, from the right,

$\lim\limits_{x \to 3^{+}} f(x) \neq f(3)$

So, we can say that the function is continuous from the left, but discontinuous from the right, at the point x = 3.

We may also have a function where it is neither continuous at a point from the left or from the right but is defined elsewhere. In that case, the graph may look like this;

NOTE:

• $f$ is continuous at a if and only if $f$ is continuous at a from the left and from the right. This can be written as:

$\lim\limits_{x \to a^{-}} f(x) \equiv f(a) \equiv \lim\limits_{x \to a^{+}} f(x)$

If you choose any x-value, you will notice that the function is continuous everywhere. This is because

$\lim\limits_{x \to a^{-}} f(x) \equiv f(a) \equiv \lim\limits_{x \to a^{+}} f(x)$

for any x.

Definition:

A function $f$ is said to be continuous on an interval if it is continuous at each and every point in the interval. Continuity at an endpoint, if it exists, means $f$ is continuous from the right (for the left endpoint) or continuous from the left (for the right endpoint).

For e.g;

(i) If $I = [a, b)$, $f$ is continuous from the right at a, and continuous at every point in the interval.

(ii) If $I = (a, b]$, $f$ is continuous from the left at b, and continuous at every other point in the interval.

(iii) If $I = [a, b]$, $f$ is continuous from the right at a and continuous from the left at b.

Theorem:

The following functions are continuous at every point of their domain.

(a) Polynomial functions.

(b) Rational functions. (The discontinuities of a rational function occur at the zeros of it’s denominator.)

(c) Exponential functions.

(d) Trigonometric functions.

(e) Logarithmic functions.

(f) Root functions.

NOTE:

• Inverse function of a continuous function is continuous.

Continuity of the algebraic combinations of functions.

If $f$ and $g$ are both continuous at x = a and c is any constant, then each of the following functions is also continuous at a:

(i) $f + g$

(ii) $f - g$

(iii) $fg$

(iv) $f/g$ if $\lim\limits_{x \to a} g(x) \neq 0$

(v) $cf$

Continuity of composite functions.

If $g$ is continuous at a, and $f$ is continuous at $g(a)$, then the composite function $(f\circ g)$ is also continuous at a.

Let’s look at an example.

Let $f(x) = x^2$, $g(x) = (x+1)$ & a = 2.

We know that

$\lim\limits_{x \to 2} g(x)$

exists here and is equal to 3. Now, we also know that $f$ is continuous at $g(2)$. And, when evaluating the limit of $f$ at $g(2)$, we have

$\lim\limits_{x \to g(2)} f(x)$ = 9

(Remember that $g(2) \equiv$ 3)

The statement says that if those two conditions above hold, then $(f\circ g)$ is also continuous at a. This means

$\lim\limits_{x \to a} f(g(x)) = f(g(a))$

Another way to put this is:

If $g$ is continuous at a, we evaluate $g(a)$ first as it is the starting function. This gives us a value back, and if the value given back is in the domain of $f$ (or, if $f$ is continuous at $g(a)$), we evaluate the function $f$ at $g(a)$.

 How clear is this post?