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

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.

Directional Continuity and Continuity over an Interval

A function $y=f(x)$ is left - continuous at $x=a$ if

$\lim _{\mathrm{x} \rightarrow \mathrm{a}^{-}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}) \quad$ or $\lim _{\mathrm{h} \rightarrow 0^{+}} \mathrm{f}(\mathrm{a}-\mathrm{h})=\mathrm{f}(\mathrm{a}) \quad$ or $\quad L H L=f(a)$
A function $y=f(x)$ is right - continuous at $x=a$ if
$\lim _{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}) \quad$ or $\quad \lim _{\mathrm{h} \rightarrow 0^{+}} \mathrm{f}(\mathrm{a}+\mathrm{h})=\mathrm{f}(\mathrm{a}) \quad$ or $\quad RHL = f(a)$

A function is said to be directional continuous at a point when the limit of the function exists and is continuous at all or specified directions.

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 Directional Continuity and Continuity Over an Interval

Solved Examples Based On 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) $(\frac{1}{2},\frac{3}{2})$

Solution:

As we have learned

Continuity in an open interval -

$\mathrm{F}(\mathrm{x})$ is said to be continuous in an open interval ( $\mathrm{a}, \mathrm{b}$ ) or $\mathrm{a}<\mathrm{x}<\mathrm{b}$ 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 $\mathrm{LHL}, \mathrm{RHL}$, 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) $f(x)=\left[x^2\right]$ 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) $\rightarrow L H L=1, f(\pi / 2)=1$;
$(B) \rightarrow L H L=1, f(2)=2$;
$(C) \rightarrow L H L=0, f(0)=0$
$(D) \rightarrow L H L=0, f(0)=0$
$f(x)=[x]$ 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:

Continuity from Right -

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

$
x=a: \text { if } \lim _{x \rightarrow a^{+}} f(x)=f(a)
$

$\ln (\mathrm{A}),(\mathrm{B})$ and $(\mathrm{D}) \rightarrow \mathrm{RHL}=$ function value
so $(A),(B),(D)$ are true
but in $(\mathrm{C}) \rightarrow \mathrm{RHL}=0, \mathrm{f}(\pi / 2)=1$
$\Rightarrow R H L \neq f(\pi / 2)$
$f(x)=[\sin x]$ 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)
$

(A),(B),(C) are the functions which are continuous at every point in $(1,3)$ and for continuity at $\lim _{\mathrm{x}=1} f(x)=f(1) \quad \lim _{x \rightarrow 1^{+}} f(x)=f(3)$ and $x \rightarrow 3^{-}$also holds true
so (A),(B),(C ) are continuous at every point of $[1,3]$
In (D), $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$ which will be discontinuous 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 $\lim _{x \rightarrow 3^{-}} f(x) \neq f(3)$
$\therefore$ discontinuous at $\mathrm{x}=2$ and $\mathrm{x}=3$