# 5.12 Continuous function  (Page 4/6)

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1. Removable discontinuity : Limit of the function exists and is finite, but is not equal to function value. We can remove this type of discontinuity by suitably redefining function value at the test point.

Problem : Find whether the given function is continuous at x = -2

| 3x – 2; x ≠ -2 f(x) = || -4 ; x = -2

Solution : Here, left and right limits, when x->2, are :

${L}_{l}={L}_{r}=L=3X-2-2=-8$

Function value at x=-2 is :

$f\left(-2\right)=-4$

Thus, function is not continuous at x=-2. The discontinuity is removable as we can remove discontinuity by redefining function, at x=-2 as f(x) = -8.

| 3x – 2; x≠ -2 f(x) = || -8 ; x = -2

2. Irremovable or jump discontinuity : This kind of discontinuity arises when left and right limits are not equal. This means limit of function does not exist.

Problem : Find whether the given function is continuous at x = 0

| |x|/x; x≠0 f(x) = || 0 ; x = 0

Solution : As a matter of fact, this is signum function. For x<0, |x| = -x, Hence, left limit is :

$\underset{x>a-}{\overset{}{\mathrm{lim}}}\frac{-x}{x}=-1$

We see that left limit is not equal to f(0) = 0. We can, therefore, conclude at this stage of analysis itself that function is not continuous at x=0. However, we continue to evaluate right hand limit as well to determine the nature of discontinuity. For x>0, |x| = x. Hence, right limit is :

$\underset{x>a+}{\overset{}{\mathrm{lim}}}\frac{x}{x}=1$

Clearly, ${L}_{l}\ne {L}_{r}$ . The discontinuity is, thus, irremovable or jump type.

3. Essential discontinuity : In this case, at least one of left or right limits does not exist or is infinite. We need to evaluate these conditions in the domain only.

Problem : Find whether the given function is continuous at x = 0.

| 1/x; x>0 F(x) = | 0 ; x = 0| -x; x<0

Solution : Here, left limit is :

$\underset{x>0-}{\overset{}{\mathrm{lim}}}f\left(x\right)=\underset{x>0-}{\overset{}{\mathrm{lim}}}x=0$

Right limit is :

$\underset{x>0-}{\overset{}{\mathrm{lim}}}f\left(x\right)=\underset{x>0-}{\overset{}{\mathrm{lim}}}\frac{1}{x}=\infty$

Since right limit is infinite, the function is discontinuous at x=0.

From these illustrations, it is clear that existence of discontinuity is associated with the manner function is defined. Here, all functions, which are discontinuous at point, are defined in piece-wise manner. On the other hand, basic functions having single definition which we have covered in the course and which are not piece wise defined are continuous functions. We do not intend to generalize these observations, but we can underline that piece - wise definitions indicate possibility of discontinuity.

Further, we note that function value exists and function is defined at the point where function is discontinuous. If there is no function value at a point, then function is not defined at that point and there is no question of continuity or discontinuity.

## Continuity in an open interval (a,b)

A function is continuous in an open interval if function is continuous at all points in the interval. This is a simple extension of the concept of continuity at a point. Both left and right limits exist and are equal to function value at all points in the interval. Since end points are not defined, there is always a point on either sides of a given point anywhere in the interval.

the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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