Continuity of Function

CONTINUITY OF A FUNCTION

A function f(x) is said to be continuous at x = a if $\underset{x\to a^{-}}{\lim }f(x)=\underset{x\to a^{+}}{\lim }f(x)$ = f(a)

i.e. L.H.L.=R.H.L. = value of the function at a i.e. $\underset{x\to a}{\lim }f(x)=f(a)$.

If f(x) is not continuous at x = a, we say that f(x) is discontinuous at x = a.

Geometrical meaning: The function 'f' will be continuous at x = a if there is no break in the graph of the function y = f(x) at the point (a, f(a)).

REASONS FOR DISCONTINUITY OF A FUNCITON

One of the following may be the reasons for the discontinuity of f(x)

(i) $\underset{x\to a^{-}}{\lim }f(x)and\underset{x\to a^{+}}{\lim }f(x)$ exist but are not equal. For example f(x)=[x] is discontinuous at all integer points.

(ii) $\underset{x\to a^{-}}{\lim }f(x)and\underset{x\to a^{+}}{\lim }f(x)$ exist and are equal but not equal to f(a). for example f(x)=[sinx] where $x\in (0,\pi )$ at $x=\frac{\pi }{2}$

(iii) when f(x) is not defined at x=a. For example f(x)=1/x-1

(iv) At least one of the limits$\underset{x\to a^{-}}{\lim }f(x)or\underset{x\to a^{+}}{\lim }f(x)$ does not exist or atleast one of these limits is $\infty or-\infty$.

CONTINUITY OF A FUNCTION IN AN INTERVAL

Continuity in an open interval:

A function f(x) is said to be continuous in an open interval (a, b) if it is continuous at each point of the interval (a, b).

Continuity in a closed interval:

A function f(x) is said to be continuous in a closed interval [a, b] if

(i) f(x) is continuous at each point of the interval (a, b).

(ii) f(x) is continuous from right at x = a i.e. $\underset{x\to a^{+}}{\lim }f(x)=f(a)$.

(iii) f(x) is continuous from left at x = b i.e.$\underset{x\to b^{-}}{\lim }f(x)=f(b)$ and;

Geometrical meaning : The function f(x) will be continuous in the closed interval [a, b] if the graph of y = f(x) is an unbroken line (curved or straight) from the point (a, f(a)) to (b, f(b)).

ALGEBRA OF CONTINUOUS FUNCTIONS

Let f(x) and g(x) be two functions, then the following results holds true.

Case I: If f and g both are continuous at x=a, then c₁f(x) $\pm$ c₂(x) and f(x) . g(x) will be continuous at x = a. And f(x)/g(x) will also be continuous at x = a, provided g(a) $\ne$ 0.

Case II: When one of the function f or g is discontinuous at x=a, then c₁f(x) $\pm$c₂(x) is definitely discontinuous, but nothing can be said about the continuity of f(x) . g(x) and f(x)/g(x). They may or may not be continuous at x=a.

Case III: When f and g both are discontinuous at x=a, then nothing can be said about the continuity of c₁f(x) $\pm$ c₂(x), f(x) . g(x) and f(x)/g(x).

CONTINUITY OF COMPOSITE FUNCTIONS

Let f(x) and g(x) are two functions, now we are interested in the continuity of f(g(x)).

Case I: If f and g both are continuous, then f(g(x)) will also be continuous.

Case II: If f is continuous and g is discontinuous. Here again two cases arises.

(a) If points of discontinuity of g(x) are not lying in the domain, the f(g(x)) will be definitely discontinuous at those points.

(b) If points of discontinuity lies in the domain, then nothing can be said about the continuity of f(g(x)) in general.

Case III: If f and g both are discontinuous, the also nothing can be about the continuity of f(g(x)).

Example 1: Find the points of discontinuity of g(f(x)) if g(x) = $\frac{1}{x^2+x-2}$ and f(x) = $\frac{1}{x-1}$.

Solution: The function f(x) = $\frac{1}{x-1}$ is discontinuous at the point x = 1.

The function g(f(x))=$\frac{1}{f^2(x)+f(x)-2}$= $\frac{1}{\left(f(x)+2\right)\left(f(x)-1\right)}$ is discontinuous at f(x) = -2 and f(x)=1.

When f(x) = -2, $\frac{1}{x-1}$=-2 $\Rightarrow$ x = $\frac{1}{2}$

When f(x) = 1, $\frac{1}{x-1}$ = 1 $\Rightarrow$ x = 2.

Hence, the composite function y = g(f(x)) is discontinuous at three points

x = 1/2, 1, 2.

TYPES OF DISCONTINUITY

Basically there are two types of discontinuity.

Removable discontinuity:

If $\underset{x\to a}{\lim }$f(x) exists but is not equal to f(a), then f(x) has a removable discontinuity at x = a and it can be removed by redefining f(x) for x = a.

Example 2: Redefine the function f(x) =[sinx] where x$\in (0,\pi )$ in such a way that it could become continuous for x $\in$(0, $\pi$).

Solution: Here $\underset{x\to \frac{\pi }{2}}{\lim }[\sin x]=0$ but $f\left(\frac{\pi }{2}\right)=1$.

Hence, f(x) has a removable discontinuity at x = $\frac{\pi }{2}$.

To remove this we redefine f(x) as follows

f(x) = [sinx], x $\in$ (0,$\pi$/2) $\cup$ ($\pi$/2,$\pi$)

= 0 , x = $\frac{\pi }{2}$.

Now, f(x) is continuous for x$\in$(0, $\pi$).

Non-removable discontinuity:

If $\underset{x\to a}{\lim }$f(x) does not exist, then we can not remove this discontinuity. So this become a non- removable discontinuity or essential discontinuity.

Example 3: Prove that f (x) = {x} has non removable discontinuity at any x$\in$I.

Solution: Since $\underset{x\to a}{\lim }f(x)$ does not exist for any a$\in$I.

Hence, f(x)= {x} has non-removable discontinuity at any x $\in$ I