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Theorems of Continuity

If we can plot the graph of a function without lifting our pen, we can say that it is a continuous function. If we lift our pen to plot a certain part of a graph, the function is discontinuous. In this article, we discuss the Theorems of Continuity. We explain the continuity of f(x)±g(x), f(x).g(x), depending upon whether f(x) and g(x) are continuous or discontinuous functions.

Continuous Function

A function, f(x), is said to be continuous at a point x = a, if and only if

  • f(a) exists
  • \(\begin{array}{l}\lim_{x\to a}f(x)=f(a)\end{array} \)
  • \(\begin{array}{l}\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=f(a)\end{array} \)
    . i.e LHL = RHL = f(a)

Theorems of Continuity are as follows.

Theorem 1

Let f(x) and g(x) be continuous functions at x = a, then

a. (f(x)+ g(x)) is continuous at x = a

b. (f(x)- g(x)) is continuous at x = a

c. (f(x). g(x)) is continuous at x = a

d. (f(x)/ g(x)) is continuous at x = a, if g(a) is not equal to zero.

If g(a) = 0, then (f(x) / g(x)) is discontinuous at x = a.

Theorem 2

If f(x) is continuous and g(x) is discontinuous at x = a, then the functions defined by

a) f(x) + g(x) will be discontinuous at x = a.

b) f(x) – g(x) will be discontinuous at x = a.

c) f(x).g(x) and f(x)/g(x), (g(a) ≠ 0) may be continuous at x = a.

Theorem 3

If g(x) is a continuous function at x = a and function f(x) is continuous at g(a), then the composition fog is continuous at x = a.

Theorem 4

If limx→ a g(x) = L and if f(x) is continuous at x = L, then limx→ a f(g(x)) = f(limx→ a g(x)) = f(L).

Also Read

Continuity of a Function

Limits Continuity and Differentiability

Solved Examples of Theorems of Continuity

Example 1:

The values of x at which function f(x) = (x-2)/(2x2+2x-4)(x4+5) is discontinuous is

a) x = 1, x = -2

b) x = -1, x = 2

c) x = -1, x = -2

d) None of the above

Solution:

Given f(x) = (x-2)/(2x2+2x-4)(x4+5)

f(x) is discontinuous => 2x2+2x-4 = 0 (Here, x4+5 will be always positive, so it cannot be zero.)

=> x2+x-2 = 0

=> (x-1)(x+2) = 0

=> x = 1, x = -2

Hence, option a is the answer.

Example 2:

Find the value of k, so that the function is continuous at x = 0.

\(\begin{array}{l}f(x)=\left\{\begin{matrix} 1-\cos 4x, & x\neq 0\\ k,& x=0 \end{matrix}\right.\end{array} \)

a) 1

b) -1

c) 0

d) None of the above

Solution:

Given

\(\begin{array}{l}f(x)=\left\{\begin{matrix} 1-\cos 4x, & x\neq 0\\ k,& x=0 \end{matrix}\right.\end{array} \)

limx→ 0f(x) = f(0) (since f(x) is continuous at x = 0)

limx→ 0f(x) = limx→ 0 1-cos 4x

= 1-1

= 0

So k = 0

Hence, option c is the answer.

Related Video

Frequently Asked Questions

Q1

Give the conditions for a function to be continuous at x = a.

A function f(x) is said to be continuous at a point x = a, if f(a) exists, lim x→a f(x) = f(a) and lim x→ a- f(x) = lim x→ a+ f(x) = f(a).

Q2

Give two examples of a continuous function.

The trigonometric functions sin x and cos x are continuous.

Q3

List two properties of continuous function.

Let f(x) and g(x) be two functions, continuous at x = a. Then, f(x) + g(x) will be continuous at x = a.
f(x) – g(x) will be continuous at x = a.

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