Continuity Over an Interval

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.

In this article, we will cover the concepts of Directional Continuity. 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.

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 $\lim _{\varepsilon \rightarrow b^{-}} f(x)=f(b)$

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 \text { 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.

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 Over an Interval

Solved Examples Based on 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 $\mathrm{x}, x^2$ 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 function is not continous 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 \text { 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 function which are continous at every point in $(1,3)$ and for coninuity at $\mathrm{x}=1$ and $\mathrm{x}=3 \lim _{x \rightarrow 1^{+}} f(x)=f(1)$ and $\lim _{x \rightarrow 3^{-}} f(x)=f(3)$ also holds true so (A),(B),(C) are continous at every pointof $[1,3]$

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

Example 4: The function $f(x)=\frac{x^3}{8}-\sin \pi x+4$ in $[-4,4]$ does not take the value
1) $-4$
2) $10$
3) $18$
4) $12$

Solution:

$
\begin{aligned}
& f(x)=\frac{x^3}{8}-\sin \pi x+4 \\
& f(-4)=-4, f(4)=12
\end{aligned}
$

$\Rightarrow \mathrm{f}(\mathrm{x})$ can take value $10$ as $\mathrm{f}(\mathrm{x})$ is continuous function

$
\frac{x^3}{8}+4 \leq 12 \text { and } I \sin \pi x \mid \leq 1 \forall x \in[-4,4]
$

$\therefore \mathrm{f}(\mathrm{x})$ cannot take the value $18$
Hence, the answer is the option (3).

Example 5: Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{g}(\mathrm{x})}{\mathrm{h}(\mathrm{x})}$, where $g$ and $h$ are continuous functions on the open interval $(\mathrm{a}, \mathrm{b})$. Which of the following statements is true for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ ?
1) $f(x)$ is continuous at all $x$ for which $x$ is not zero.
2) $f(x)$ is continuous at all $x$ for which $g(x)=0$.
3) $f(x)$ is continuous at all $x$ for which $g(x) \neq 0$
4) $f(x)$ is continuous at all $x$ for which $h(x)$ is not equal to zero.

Solution:

$
f(x)=\frac{g(x)}{h(x)}
$

By theorem, if $g$ and $h$ are continuous functions on the open interval $(\mathrm{a}, \mathrm{b})$, then $\frac{g}{\mathrm{~h}}$ is also continuous at all $x$ in the open interval $(\mathrm{a}, \mathrm{b})$, where $\mathrm{h}(1)$ is not equal to zero. Hence, the answer is the option (4).

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.