Directional Continuity and Continuity over an Interval - Practice Questions & MCQ

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

$
\lim _{\mathrm{x} \rightarrow \mathrm{a}^{-}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a}) \quad \text { or } \quad \lim _{\mathrm{h} \rightarrow 0^{+}} \mathrm{f}(\mathrm{a}-\mathrm{h})=\mathrm{f}(\mathrm{a}) \quad \text { 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 \text { or } \quad \lim _{\mathrm{h} \rightarrow 0^{+}} \mathrm{f}(\mathrm{a}+\mathrm{h})=\mathrm{f}(\mathrm{a}) \quad \text { or } \quad R H L=f(a)
$
For example,
$\mathrm{f}(\mathrm{x})=\mathrm{y}=[\mathrm{x}], \quad($ where, [.] is G.I.F $)$. is right continuous at $\mathrm{x}=2$

$
f(2)=\lim _{x \rightarrow 2^{+}}[x]=2
$
But it is NOT left continuous at $x=2$ as $L H L=1$ but $f(2)=2$. Hence $f(2)$ does not equal $L H L$ 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 $c \in(a, b), f(x)$ 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 $x=a$

And 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 $\mathrm{x}=\mathrm{b}$

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

$
\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 the left neighbourhood 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 third condition is not satisfies and hence $f(x)$ is not continuous in $[2,3]$

Discontinuity and Removable Types Discontinuity

A non-continuous function 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 $\mathrm{x}=\mathrm{a}$ is equal to $\mathrm{f}(\mathrm{a})$, i.e. $\lim _{x \rightarrow a} f(x)=f(a)$.

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

Consider the function $f(x)=\frac{x^2-4}{x-2}$, where, $x \neq 2$

$
\therefore \quad f(x)=\frac{(x+2)(x-2)}{(x-2)}=(x+2), x \neq 2
$

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

Here $\mathrm{LHL}=\mathrm{RHL}=4$
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{array}{cc}
\frac{x^2-4}{x-2}, & x \neq 2 \\
4, & x=2
\end{array}\right.
$
It makes the function continuous as now LHL $=$ RHL $=f(2)$