Continuity of a Function

In this section we will deal with the concept of continuity, one of the most important and also one of the most fascinating ideas in all of mathematics.

Although the meaning of the word continuous seems intuitively clear to most student, it is not obvious how a good definition of this idea should be formulated. One popular dictionary explains continuity as follows:

Continuity: Quality or state of being continuous.

Continuous: Having continuity of parts.

Trying to learn the meaning of continuity from these two statements alone is like trying to learn Latin with only a Latin dictionary.

Before we present the precise definition of continuity, we shall briefly discuss the concept in an informal and intuitive way to give the students a feeling for its meaning.

Roughly speaking, the situation is this: Suppose a function f has the value f(a) at a certain point a , then f is said to be continuous at ‘a’ if at every nearby point x the function value f(x) is close to f (a). Another way of putting it is as follows:

If we let x move towards a, we want the corresponding function value f(x) to become arbitrarily close to f (a), regardless of the manner in which x approaches a , we do not want sudden jumps in the value of a continuous function

Definition of continuity:

A function f(x) is said to be continuous at x = a if

$\large \lim_{x\rightarrow a^-}f(x) = \lim_{x\rightarrow a^+}f(x) = f(a) $

i.e. L.H.L.=R.H.L. = value of the function at a i.e. $\large \lim_{x\rightarrow a}f(x) = f(a)$

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

f(x) will be discontinuous at x = a in any of the following cases:

(i) f(a) is not defined.

(ii) At least one of the limits does not exist.

(iii) $\large \lim_{x\rightarrow a^-}f(x) $ and  $\large \lim_{x\rightarrow a^+}f(x) $  exist but are not equal.

(iv)$\large \lim_{x\rightarrow a^-}f(x) $ and  $\large \lim_{x\rightarrow a^+}f(x) $   exist and are equal but not equal to f(a).

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 (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 from right at x = a i.e. $\large \lim_{x\rightarrow a^+}f(x) = f(a) $

(ii) f(x) is continuous from left at x = b i.e. $\large \lim_{x\rightarrow b^-}f(x) = f(b) $ and

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

Illustration : Let f(x) be a continuous function and g(x) be a discontinuous function. Prove that f(x)+ g(x) is a discontinuous function.

Solution: Suppose that h(x) = f(x) + g(x) is continuous.

Then in view of the fact that f(x) is continuous . g(x) = h(x) − f(x),

as difference of continuous functions, is continuous.

But this is a contradiction, since g(x) is given as a discontinuous function.

Hence, f(x) + g(x) is discontinuous.

Illustration : Discuss the continuity of

$ \large f(x) = \left\{\begin{array}{llll} |x + 1| \; , x < -2 \\ 2x+3 \; , -2 \le x < 0  \\ x^2+3 \; , 0 \le x < 3 \\ x^3-15 \; ,  x \ge 3\end{array} \right. $

Solution: We write f(x) as

$ \large f(x) = \left\{\begin{array}{llll} |x + 1| \; , x < -2 \\ 2x+3 \; , -2 \le x < 0  \\ x^2+3 \; , 0 \le x < 3 \\ x^3-15 \; ,  x \ge 3\end{array} \right. $

As we can see, f(x) is defined as a polynomial function in each of intervals

(− ∞, −2), (−2 , 0), (0 , 3) and (3, ∞).

Therefore, it is continuous in each of these four open intervals. Thus we check the continuity at

x = −2 , 0, 3.

At the point x = −2

limx−>−2 f(x) = limx−>−2 (−x − 1) = + 2 − 1 = 1

limx−>−2+ f(x) = limx−>−2+ (2x + 3) = 2 . (−2) + 3 = −1

Therefore, limx−>−2 f(x) does not exist and hence f(x) is discontinuous at x = −2.

At the point x = 0

limx−>0 f(x) = limx−>0 (2x + 3) = 3

limx−>0+ f(x)= limx->0+(x2 + 3) = 3

f(0) = 02 + 3 = 3

Therefore, f(x) is continuous at x = 0.

At the Point x = 3.

limx−>3 f(x) = limx−>3 (x2 + 3) = 32 + 3 = 12

limx−>3+ f(x) = limx−>3+ (x3 − 15) = 33 − 15 = 12

f(3) = 33 − 15 = 12

Therefore, f(x) is continuous at x = 3

We find that f(x) is continuous at all points in R except at x = −2

Also Read

→Limits of a Function
→Continuity of a Function
→Differentiability of a Function
→Rules for differentiation
→Higher order derivatives
→L’HOSPITAL’S RULE

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