Continuity over an Interval

Q: What is continuity from right?

The function f(x) is said to be continuous from right at <img alt="x=a:if\:\lim_{x\rightarrow a^{+}}\:f(x)=f(a)" data-cke-widget-data="%7B%22hasCaption%22%3Afalse%2C%22src%22%3A%22https%3A%2F%2Fentrancecorner.oncodecogs.com%2Fgif.latex%3Fx%253Da%253Aif%255C%253A%255Clim_%257Bx%255Crightarrow%2520a%255E%257B%26plus%3B%257D%257D%255C%253Af%2528x%2529%253Df%2528a%2529%22%2C%22alt%22%3A%22x%3Da%3Aif%5C%5C%3A%5C%5Clim_%7Bx%5C%5Crightarrow%20a%5E%7B%2B%7D%7D%5C%5C%3Af(x)%3Df(a)%22%2C%22width%22%3A%22%22%2C%22height%22%3A%22%22%2C%22lock%22%3Atrue%2C%22align%22%3A%22none%22%2C%22classes%22%3Anull%7D" data-cke-widget-upcasted="1" src="https://cache.careers360.mobi/media/uploads/froala_editor/images/1725888157032.png" class="fr-fic fr-dii">

Q: What is continuity from the left?

The function f(x) is said to be continuous from the left at x = a if <img alt="\lim_{x\rightarrow a^{-}}\:f(x)=f(a)" data-cke-widget-data="%7B%22hasCaption%22%3Afalse%2C%22src%22%3A%22https%3A%2F%2Fentrancecorner.oncodecogs.com%2Fgif.latex%3F%255Clim_%257Bx%255Crightarrow%2520a%255E%257B-%257D%257D%255C%253Af%2528x%2529%253Df%2528a%2529%22%2C%22alt%22%3A%22%5C%5Clim_%7Bx%5C%5Crightarrow%20a%5E%7B-%7D%7D%5C%5C%3Af(x)%3Df(a)%22%2C%22width%22%3A%22%22%2C%22height%22%3A%22%22%2C%22lock%22%3Atrue%2C%22align%22%3A%22none%22%2C%22classes%22%3Anull%7D" data-cke-widget-upcasted="1" src="https://cache.careers360.mobi/media/uploads/froala_editor/images/1725888183612.png" class="fr-fic fr-dii"> .

Q: What is Continuity in an open interval?

f(x) is said to be continuous in an open interval (a, b) or a < x < b. If it is continuous at every point of the interval belonging to (a, b).

Continuity Over an Interval

Edited By Komal Miglani | Updated on Sep 09, 2024 06:53 PM IST

Continuity and Discontinuity is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which graphs of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of Continuity and Discontinuity have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concepts of Directional Continuity and Discontinuity. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including one in 2021, and two in 2023.

Directional Continuity and Continuity over an Interval

Continuity: A real function f is said to be continuous if it is continuous at every point in the domain of f.

This definition requires a bit of elaboration. Suppose f is a function defined on a closed interval [a, b], then for f to be continuous, it needs to be continuous at every point in [a, b] including the end points a and b. Continuity of f at a means $\lim_{x\rightarrow a^{+}}f(x)=f(a)$

and continuity of f at b means $\lim_{x\rightarrow b^{-}}f(x)=f(b)$

Observe that $\lim_{x\rightarrow a^{-}}f(x)$ and $\lim_{x\rightarrow b^{+}}f(x)$ do not make sense. As a consequence
of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of f is a singleton, f is a continuous function.

Geometrical interpretation of continuity at a point

When a graph breaks at a particular point when it approaches from left and right.

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

So limit exists but is not continuous: but when it is equal to f(a) at x = a then f(x) is continuous.

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

Continuity in a closed interval

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

$a\leq x\leq b$ if

1. f is continuous at each and every point in ( a, b)

2. Right hand limit at x = a must exist and

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

3. Left hand limit at x = b must exist and

$\lim_{x\rightarrow b^{-}}\:f(x)=f(b)$

So continuity can be defined in two ways: Continuity at a point and Continuity over an interval.

Directional Continuity and Continuity over an Interval

A function may happen to be continuous in only one direction, either from the “left” or from the “right”.

A function y = f(x) is left - continuous at x = a if $\\\mathrm{\lim_{x\rightarrow a^-}\;f(x)=f(a)\;\;\;\;or\;\;\;\lim_{h\rightarrow 0^+}\;f(a-h)=f(a)}\;\;\;\;or \;\;\;\;LHL=f(a)$

A function y = f(x) is right - continuous at x = a if $\\\mathrm{\lim_{x\rightarrow a^+}\;f(x)=f(a)\;\;\;\;or\;\;\;\lim_{h\rightarrow 0^+}\;f(a+h)=f(a)}\;\;\;\;or\;\;\;\;RHL=f(a)$

For example,

$\\\mathrm{f(x)=y=[x],\;\;\;\left (where,[.]\;is\;G.I.F \right ).\;\;is \;right\;continuous\;at\;x=2}\\\\f(2)=\lim_{x\rightarrow 2^+}[x]=2$

But it is NOT left continuous at x = 2 as LHL = 1 but f(2) = 2. Hence f(2) does not equal LHL at x = 2.

Continuity over an Interval

Over an open interval (a, b)

A function f(x) is continuous over an open interval (a, b) if f(x) is continuous at every point in the interval.

For any c ∈ (a, b), f(x) is continuous if

$\lim_{x\rightarrow c^-}\;f(x)=\lim_{x\rightarrow c^+}\;f(x)=f(c)$

Over a closed interval [a, b]

A function f(x) is continuous over a closed interval of the form [a, b] if

it is continuous at every point in (a, b) and
is right-continuous at x = a and
is left-continuous at x = b.

i.e.At x = a, we need to check $f(a)=\lim _{x \rightarrow a^{+}} f(x)\,\,\,(=\lim _{h \rightarrow 0^+} f(a+h)=\mathrm{R.H.L.})$ . L.H.L. should not be evaluated to check continuity x = a

And at x = b, we need to check $f(b)=\lim _{x \rightarrow b^{-}} f(x) \,\,\,(=\lim _{h \rightarrow 0^+} f(b-h)=\mathrm{L.H.L.})$ . R.H.L. should not be evaluated to check continuity x = b

Consider one example,

f(x) = [ x ], prove that this function is not continuous in [2, 3],

Sol.

Condition 1

For continuity in (2,3)

At any point x = c lying in (2,3),

f(c) = [c] = 2 ( as c lies in (2,3))

LHL at x = c : $\lim _{x \rightarrow c^{-}} [x]=2$ (as in close left neighbourhood of x = c, the function equals 2)

RHL at x = c: $\lim _{x \rightarrow c^{+}} [x]=2$ (as in close right neighbourhood of x = c, the function equals 2)

So function is continuous for any c lying in (2,3). Hence the function is continuous in (2,3)

Condition 2

Right continuity at x = 2

f(2) = 2

$\\\mathrm{\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2^+}[x]=\lim_{h\rightarrow 0^+}[2+h]=2}$

So f(x) is left continuous at x = 2

Condition 3

Left continuity at x = 3

f(3) = 3 and

$\lim_{x\rightarrow 3^-}[x]=\lim_{h\rightarrow 0^+}[3-h]=2$

(as in left neghbourhood of 3, f(x) = 2)

So f(3) does not equal LHL at x = 3

hence f(x) is not left continuous at x = 3

So the third condition is not satisfied and hence f(x) is not continuous in [2,3]

Discontinuity and Removable Types Discontinuity

A function that is non-continuous, is said to be a discontinuous function.

There are various kinds of discontinuity at a point, which are classified as shown below:

1. Removable Discontinuity

2. Non-Removable Discontinuity

Finite Type
Infinite Type
Oscillatory

Removable Discontinuity

In this type of discontinuity, the limit of the function $\lim_{x\rightarrow a}\;f(x)$ necessarily exists but it is either not equal to f(a) or f(a) is not defined. However, it is possible to redefine the function at x = a in such a way that the limit of the function at x = a is equal to f(a), i.e. $\lim_{x\rightarrow a}\;f(x)=f(a)$ .

Again removable discontinuity can be classified into missing point discontinuity and isolated point discontinuity

$\\\text{Consider the function }f(x)=\frac{x^2-4}{x-2},\;\;\text{where, }x\neq2\\\therefore \;\;\;\;\;\;f(x)=\frac{(x+2)(x-2)}{(x-2)}=(x+2),\;x\neq2$

In the above graph, observe that the graph has a hole (missing point) at x = 2, which makes it discontinuous at x = 2.

Here LHL = RHL = 4

Here function f(x) is not defined at x = 2, i.e. f(2) is not defined. However, we can redefine the function as

$f(x)=\left\{\begin{matrix} \frac{x^2-4}{x-2}, & \;\;x\neq2\\ 4, & \;\;x=2 \end{matrix}\right.$

It makes the function continuous as now LHL = RHL = f(2)

Solved Examples Based On the Directional Continuity and Continuity over an Interval:

Example 1: $f(x)= [x]$ is continuous at each point of which of the following intervals?

1) (1,2)

2) (1,3)

3) (-1,1)

4) (1/2,3/2)

Solution:

As we have learned

Continuity in an open interval -

F(x) is said to be continuous in an open interval (a, b) or a < x < b if it is continuous at each and every point of the interval belonging to its domain.

f(x)=[x] will be discontinuous at integers . In (B), (C), (D) there are integers, lying in the interval in (B), (C), and (D), f(x) will be continuous at each point. But in (A) it is

Hence, the answer is the option 1.

Example 2: $f(x)=x^{2}$ is continuous at each point of which of the following intervals?

1) (1,5)

2) (5,7)

3) (7,9)

4) All of them

Solution:

As we have learned

Continuity in an open interval -

F(x) is said to be continuous in an open interval (a, b) or a < x < b. If it is continuous at each and every point of the interval belonging to its domain.

At every x, $x^{2}$ will give LHL, RHL, and function value all three equal so continuous everywhere so in all intervals it will be continuous

Hence, the answer is the option 1

Example 3: Which of the following statements is false?

1) $f(x)= \sin x$ is left continuously at $x= \pi /2$

2) $f(x)= [ x]$ is left continuous at $x= 2$

3) $f(x)= |x|$ is left continuous at $x= 0$

4) $f(x)= [x^{2}]$ is left continuous at $x= 0$

Solution: To check left continuity we need to find LHL and function value at the point x = a

$(A)\rightarrow LHL =1, f(\pi /2)=1;$

$(B)\rightarrow LHL =1, f(2)=2;$

$(C)\rightarrow LHL =0, f(0)=0;$

$(D)\rightarrow LHL =0, f(0)=0$

f(x)=[x] is not left-continuous at x=2

Hence, the answer is the option 2.

Example 4: Which of the following statements is false? ([.]= G.I.F)

1) $f(x)= [x]$ is continuous from right at $x=2$

2) $f(x)= [\sin x]$ is continuous from right at $x=-\pi/2$

3) $f(x)= [\sin x]$ is continuous from right at $x=\pi/2$

4) $f(x)= x^{2}$ is continuous from right at $x=2$

Solution:

As we have learned

Continuity from Right -

The function f(x) is said to be continuous from right at

$x=a:if\:\lim_{x\rightarrow a^{+}}\:f(x)=f(a)$

In (A),(B) and (D) $\rightarrow$ RHL = function value

so (A),(B),(D) are true

but in (C) $\rightarrow$ RHL =0 ,f( $\pi /2$ ) =1

$\Rightarrow RHL \neq f(\pi /2)$

f(x) =[sin x] is not continuous from right at x = $\pi /2$

Hence, the answer is the option 3.

Example 5: Which of the following functions is not continuous at all x being in the interval [1,3]?

1) $f(x)= x^{2}$

2) $f(x)= x^{3}$

3) $f(x)= \sin x$

4) $f(x)= [x]$

Solution:

As we have learned

Continuity from Right -

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

$a\leq x\leq b$ if

1. f is continuous at each and every point in ( a, b)

2. Right hand limit at x = a must exist and

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

3. Left hand limit at x = b must exist and

$\lim_{x\rightarrow b^{-}}\:f(x)=f(b)$

- wherein

(A),(B),(C) are the functions which are continuous at every point in (1,3) and for continuity at x=1 and x= 3 $\lim_{x\rightarrow 1^{+}}f(x)= f(1)$ and $\lim_{x\rightarrow 3^{-}}f(x)= f(3)$ also holds true

so (A),(B),(C ) are continuous at every point of [1,3]

In (D), f(x) =[x] which will be discontinuous at x=2 and x=3 both as $\lim_{x\rightarrow 2^{+}}f(x)$ , $\lim_{x\rightarrow 2^{-}}f(x)$ and f(2) are not all equal and $\lim_{x\rightarrow 3^{-}}f(x)\neq f(3)$

$\therefore$ discontinuous at x= 2 and x= 3

Summary

Continuity and Discontinuity is an important concept in Calculus. It provides a deep understanding of how the functions interact and change. It is very helpful in practical applications for physics, economics, etc. There are various kinds of discontinuity at a point. The graph shows the change in the function maximum and minimum. Overall, this provides a better interpretation of the functions leading to more accurate solutions.