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)

Right

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)

Left

Here, it is clear from the graph that the function is continuous from the left as x 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;

NotContinuous

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)

Continuous

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?