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Problem : Obtain the continuous extension of the function given by :

f x = x 2 1 x 1 ; x 1

Solution : The denominator is zero when x=1. In order that this point is included in the domain of the function, the value of function at this point should be equal to the limit of the function at this point. Now,

lim x 1 x 2 1 x 1 = x + 1 = 2

Hence, extended function is :

f x = x 2 1 x 1 ; x 1

= 2 ; x = 1

Combination of continuous and discontinuous functions

We can have combinations of function resulting from function operations or composition, which involves both continuous and discontinuous functions. We need to know what would be the nature of resulting function? In general, such combinations result in discontinuous function. It is not important to have a generalization here, but such combinations point to the possibility that a function may have discontinuities.

Let us consider two functions f(x) and g(x). Let also assume that f(x) is a continuous function and g(x) is discontinuous function. We, now, consider the operation as :

h x = f x + g x

Rearranging, we have :

g x = h x f x

In order to test the nature of h(x), let us assume that h(x) is a continuous function. In that case, we know that difference of two continuous functions is a continuous function. As such, g(x) is a continuous function. But, this is contradictory to what we had assumed in the beginning. It means that our supposition that h(x) is continuous function, is wrong. Clearly, function h(x) resulting from addition operation is a discontinuous function. We can extend similar conclusion to subtraction operation as well as subtraction is equivalent to addition operation.

We have already studied few discontinuous functions like greatest integer, least integer and fraction part functions etc. They are discontinuous at integral values. A composition such as y = sin[x] is a discontinuous function. We can analyze these compositions analytically. However, graphical methods are more efficient in this case. We can draw these compositions with the help of transformation of graph by these discontinuous functions. We have dealt this aspect separately in modules titled such as “transformation of graphs by greatest integer function” or “transformation of graphs by fraction part function” etc. For this reason, we shall not discuss this topic any further here in this module.

Examples

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

| g(x); x ≠ 1 f(x) = || 2 ; x = 1

where,

g x = x 3 1 x 2 1

Solution : In order to factorize cubic expression in the numerator, we guess that x=1 is one real root. We can see that expression turns zero for x=0. Hence, we conclude that (x-1) is one of the factor of cubic expression. Now, for x ≠ 1,

f x = x 3 1 x 2 1 = x 1 x 2 + x + 1 x 1 = x 2 + x + 1 x + 1

The reduced expression is not an indeterminate form. Hence, left and right limits, when x->1, are :

L l = L r = L = 3 2

Here, f 1 = 2 . We, therefore, conclude that function is not continuous at x=1. This is a removable discontinuity as we can remove this discontinuity by redefining function at x=1 as f(x) = 3/2.

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

| x sin(1/x), x ≠ 0 f(x) = || 0; x = 0

Solution : For x ≠ 0;

lim x > 0 x sin 1 x = 0

Note that as x-->0, 1/x-->infinity and sin(1/x) -->a finite value in [-1,1]. Thus, function tends to become 0 X finite value, which is equal to zero. Similar is case for right limit. Hence,

L l = L r = L = f 0 = 0

Thus, function is continuous at x = 0.

Exercises

If the function given below is continuous in its domain, then determine value of k.

f x = x 3 + x 2 3 x + 3 x 1 2 ; x 1

= k ; x = 1

For x ≠1, the function is :

f x = x 3 + x 2 3 x + 3 x 1 2

We guess here that one of the roots of numerator is 1. It is true as numerator is zero for x=1. Thus, (x-1) is a factor of cubic expression in the numerator.

f x = x 1 x 2 + 2 x 3 x 1 2 = = x 1 x 1 x + 3 x 1 2

f x = x + 3

Clearly,

lim x 1 f x = lim x 1 x + 3 = 4

For function to be continuous, this limit should be equal to value of function at x=1. Hence,

f 1 = k = 4 k = 4

Determine if the given function is continuous.

f x = x sin x ; x 0

= 0 ; x = 0

For x ≠ 0, the function is of standard form whose limit is 1 when x approaches 0. Thus, limit is not equal to function value at x=0. Clearly, function is discontinuous at x=0. It is a removable discontinuity.

Determine continuity of function given at x= 0 :

f x = e 1 / x 1 + e 1 / x ; x 0

= 0 ; x = 0

In order to evaluate limit at x=0, we determine left and right limits at x=0. We see here that as x approaches zero 1/x approaches negative infinity from left, whereas it approaches to positive infinity from the right. Using these facts, left hand limit is :

lim x 0 e 1 + e = 0 1 + 0 = 0

Now right limit is :

lim x 0 + e 1 / x 1 + e 1 / x [ form ]

Dividing by e 1 / x , we have :

lim x 0 + 1 1 e 1 / x + 1 = 1 1 e + 1 = 1

Clearly, L l L r . Hence, function is discontinuous at x=0.

Determine continuity of the function given by :

f x = x ; x is rational

= 2 x ; x is irrational

In order to determine continuity, we consider set of rational and irrational numbers separately. Let us first work with rational set. We determine limit when as x approaches any real value point “a” in real domain. Note that x approaches real number “a” assuming only rational number in its set. We say that x approaches "a" through rational numbers. Then,

lim x a f x = lim x a x = a

Now, we work with irrational set. Here, x approaches real number “a” assuming only irrational numbers in the domain. Then,

lim x a f x = lim x a 2 x = 2 a

Let us consider that function is continuous at x=a. In that case,

L l = L r a = 2 a a = 1

We have taken any arbitrary point x=a and we find that function is continuous only for a single value – not an interval or union of intervals. Thus, we conclude given function is continuous only at one point and discontinuous at all other points.

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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