Continuity and Discontinuity: Definition, Examples, Questions

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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This Story also Contains

1. Continuity and Discontinuity
2. Solved Examples Based on Continuity and Discontinuity
3. Summary

In this article, we will cover the concepts of 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 twenty-six questions have been asked on this concept, including one in 2013, one in 2015, one in 2016, one in 2018, four in 2019, two in 2020, nine in 2021, five in 2022, and two in 2023.

Continuity and Discontinuity

Suppose $f$ is a real function on a subset of the real numbers and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if

$
\lim _{x \rightarrow c} f(x)=f(c)
$

More elaborately, if the left-hand limit, right-hand limit, and the value of the function at $x=c$ exist and are equal to each other, then $f$ is said to be continuous at $x=c$.

Recall that if the right-hand and left-hand limits at $x=c$ coincide, then we say that the common value is the limit of the function at $\mathrm{x}=\mathrm{c}$.

Hence we may also rephrase the definition of continuity as follows:

A function is continuous at $\mathrm{x}=\mathrm{c}$ if the function is defined at $x=c$ and if the value of the function at $x=c$ equals the limit of the function at $x=c$.

If $f$ is not continuous at $c$, we say $f$ is discontinuous at $c$, and $c$ is called a point of discontinuity of $f$.

If the function is continuous, Its graph does not break but for discontinuous functions, there is a break in the graph. A real function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper. In case one has to lift the pen at a point, the graph of the function is said to have a break or discontinuity at that point, say $x=a$.

Many functions have the property that their graphs does not have a break. Such functions are called continuous functions. Other functions have points at which a break in the graph occurs, but satisfy this property over some intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs.

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

Continuity at a point

Let us see different types of conditions to see continuity at point $x = a$

We see that the graph of $f(x)$ has a hole at $x=a$, which means that $f(a)$ is undefined. At the very least, for $f(x)$ to be continuous at $x=a$, we need the following conditions:

(i) $f(a)$ is defined

Next, for the graph given below, although $f(a)$ is defined, the function has a gap at $x=a$. In this graph, the gap exists because lim $\lim\limits _{x → a }f(x)$ does not exist. We must add another condition for continuity at $x=a$, which is

(ii) $\lim\limits _{x \rightarrow a} f(x)$ exists

The above two conditions by themselves do not guarantee continuity at a point. The function in the figure given below satisfies both of our first two conditions but is still not continuous at $a$. We must add a third condition to our list:

(iii) $\lim\limits _{x \rightarrow a} f(x)=f(a)$

So, a function $f(x)$ is continuous at a point $x = a$ if and only if the following three conditions are satisfied:

i) $f(a)$ is defined

ii) $\lim\limits_{x \rightarrow a} f(x)$ exists

iii) $\lim\limits_{x \rightarrow a} f(x)=f(a)$ or

$\begin{aligned} & \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a) \\ & \text { i.e. } \text { L.H.L. }=\text { R.H.L. }=\text { value of the function at } x=a\end{aligned}$

A function is discontinuous at a point $a$ if it fails to be continuous at $a$.

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 $c \in(a, b), f(x)$ 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 $\mathrm{x}=\mathrm{a}$, we need to check $f(a)=\lim _{x \rightarrow a^{+}} f(x)\left(=\lim _{h \rightarrow 0^{+}} f(a+h)=\right.$ R.H.L. $)$.

L.H.L. should not be evaluated to check continuity of the first element of the interval, $x=a$

Similarly, at $\mathrm{x}=\mathrm{b}$, we need to check $f(b)=\lim _{x \rightarrow b^{-}} f(x)\left(=\lim _{h \rightarrow 0^{+}} f(b-h)=\right.$ L.H.L. $)$.

R.H.L. should not be evaluated to check continuity of the last element of the interval $x=b$

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

RHL at $\mathrm{x}=\mathrm{c}: x \rightarrow \mathrm{c}^{+}[x]=2$ (as in close right neighbourhood of $\mathrm{x}=\mathrm{c}$, 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 $x=2$

$
\begin{aligned}
& f(2)=2 \\
& \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}[x]=\lim _{h \rightarrow 0^{+}}[2+h]=2
\end{aligned}
$

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 $[2,3]$

Recommended Video Based on Continuity and Discontinuity

Solved Examples Based on Continuity and Discontinuity

Example 1: Let $a, b \in \mathbf{R},(a \neq 0)$. If the function is defined as $
f(x)= \begin{cases}\frac{2 x^2}{a}, & 0 \leq x<1 \\ a, & 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^3}, & \sqrt{2} \leq x<\infty\end{cases}
$ is continuous in the interval $[0, \infty)$, then an ordered pair $(a, b)$ is :

1) $(\sqrt{2}, 1-\sqrt{3})$
2) $(-\sqrt{2}, 1+\sqrt{3})$
3) $(\sqrt{2},-1+\sqrt{3})$
4) $(-\sqrt{2}, 1-\sqrt{3})$

Solution:

The given function is

$f(x)=\left\{\begin{array}{ccc}\frac{2 x^2}{a} & & 0 \leq x<1 \\ a & & 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^3}, & & \sqrt{2} \leq x<\infty\end{array}\right.$

since the function is continuous
$
\begin{aligned}
& \Rightarrow \text { continuous at } x=1 \text { and } x=\sqrt{2} \\
& \operatorname{Lim}_{x \rightarrow 1^{-}} f(x)=\operatorname{Lim}_{x \rightarrow 1^{+}} f(x)=f(1) \\
& \Rightarrow \frac{2}{a}=a \Rightarrow a^2=2
\end{aligned}
$
$
\begin{aligned}
& \text { and } \operatorname{Lim}_{x \rightarrow \sqrt{2}} f(x)=\operatorname{Lim}_{x \rightarrow \sqrt{2}} f(x)=f(\sqrt{2}) \\
& \Rightarrow a=\frac{2 b^2-4 b}{2 \sqrt{2}} \\
& \Rightarrow b^2-2 b=\sqrt{2} a
\end{aligned}
$
If $a=\sqrt{2}$ then $b^2-2 b-2=0 \Rightarrow b=1 \pm \sqrt{3}$
If $a=-\sqrt{2}$ then $b^2-2 b+2=0 \Rightarrow b$ is imaginary which is not possible $\Rightarrow(a, b)=(\sqrt{2}, 1+\sqrt{3})$ or $(\sqrt{2}, 1-\sqrt{3})$

Example 2: If the function $
g(x)= \begin{cases}k \sqrt{x+1}, & 0 \leq x \leq 3 \\ \mathrm{~m} x+2, & 3<x \leq 5\end{cases}
$ is differentiable, then the value of $k+m$ is :

1) $2$
2) $\frac{16}{5}$
3) $\frac{10}{3}$
4) $4$

Solution:

As we have learned

Continuity at a point -

A function $f(x)$ is said to be continuous at $x = a$ in its domain if

1. $f(a)$ is defined : at $x=a$.
2. $\lim\limits _{x \rightarrow a} f(x)$ exists means $L H L=R H L$
3. $\lim \limits_{x \rightarrow a} f(x)=f(a)$ then the limit equals the value at $x=a$

Condition for differentiability -

A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

For continuity at $\mathrm{x}=3$,

$
k \times 2=3 m+2 \text {. }
$

and for diffrentiability at $x=3$,

$
\begin{aligned}
& \left.\frac{k}{2 \sqrt{x+1}}\right|_{x=3}=m \\
& \Rightarrow K=4 m \ldots \ldots(2)
\end{aligned}
$

solving ( 1 ) and (2)

$
m=2 / 5
$

and $\mathrm{k}=8 / 5$
therefore $\mathrm{m}+\mathrm{k}=2$

Example 3: Consider the function : $f(x)=[x]+[1-x]$, $-1 \leq x \leq 3$ when $[x]$ is the greatest integer function.

Statement I: $f$ is not continuous at $\mathrm{x}=0,1,2$ and $3$ .
Statement II: $\quad f(x)=\left\{\begin{array}{cc}-x & -1 \leq x<0 \\ 1-x & 0 \leq x<1 \\ 1+x & 1 \leq x<2 \\ 2+x & 2 \leq x \leq 3\end{array}\right.$

1) Statement I is true; Statement II is false.

2) Statement I is true; Statement II is true;

Statement II is not a correct explanation for Statement I.

3) Statement I is true; Statement II is true;

Statement II is a correct explanation for Statement I.

4) Statement I is false; Statement II is true

Solution:

Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]+[1-\mathrm{x}],-1 \leq x \leq 3$
where $[\mathrm{x}]$ is the greatest integer function.
$f$ is not continuous at $x=0,1,2,3$
But in statement-2 $f(x)$ is continuous at $x=3$
Hence, statement- 1 is true and 2 is false.
Hence, the answer is option 1.

Example 4: $
f(x)=\left\{\begin{array}{c}
(x-1)^{\frac{1}{2-x}}, x>1, x \neq 2 \\
k, x=2
\end{array}\right.
$

of $k$ for which $f$ is continuous at $x=2$ is:
1) $1$
2) $e$
3) $e^{-1}$
4) $e^{-2}$

Solution:

As we learned,

The rule for a continuous function:

A function is continuous at $x=a$ if and only if

$
L=R=V
$

L.H.L $=$ R.H.L $=$ value at $x=a$.
where

$
\begin{aligned}
& L=\lim _{x \rightarrow a^{-}} f(x) \\
& R=\lim _{x \rightarrow a^{+}} f(x) \\
& V_I=\lim _{x \rightarrow a} f(x)
\end{aligned}
$

$
\begin{aligned}
& \text { for } x=2^{+}
\end{aligned}
$

Limit

$
\begin{aligned}
& \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=\lim _{x \rightarrow 2}(1+x-2)^{\frac{1}{2-x}} \\
& \lim _{x \rightarrow 2}=e^{\frac{x-2}{2-x}}=e^{-1}
\end{aligned}
$

Thus, $k=e^{-1}$
Hence, the answer is the option 3.

Example 5:

Let $f: R \rightarrow R_{\text {be a function defined as }}$

$
f(x)= \begin{cases}\frac{\sin (a+1) x+\sin 2 x}{2 x} & , \text { if } x<0 \\ b, \text { if } x=0 \\ \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^5 / 2} & \text { if } x>0\end{cases}
$

If continuous at $x=0$, then the value of $a+b$ is equal to:
1) $-\frac{3}{2}$
2) -3
3) $-\frac{5}{2}$
4) -2

Solution:
$f(x)$ is continuous at $x=0$

$
\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x) \\
& f(0)=\mathrm{b}
\end{aligned}
$

$
\begin{aligned}
\lim _{x \rightarrow 0^{-}} f(x) & =\lim _{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right) \\
& =\frac{a+1}{2}+1 \\
\lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}} \\
& =\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^3-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^3}+\sqrt{x}\right)} \\
& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^2}+1\right)}=\frac{1}{2}
\end{aligned}
$

from (2), (3) and (4),(1)

$
\begin{aligned}
& \frac{1}{2}=\mathrm{b}=\frac{\mathrm{a}+1}{2}+1 \\
& \Rightarrow \mathrm{~b}=\frac{1}{2}, \mathrm{a}=-2 \\
& \mathrm{a}+\mathrm{b}=\frac{-3}{2}
\end{aligned}
$

Hence, the answer is option 1 .

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 better interpretation of the functions leading to more accurate solutions.