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

$\underset{\mathrm{x}\to {\mathrm{a}}^{}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})\text{ or }\underset{\mathrm{h}\to 0^{+}}{lim}\mathrm{f}(\mathrm{a}\mathrm{h})=\mathrm{f}(\mathrm{a})\text{ or }LHL=f(a)$
A function $y=f(x)$ is right continuous at $x=a$ if

$\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})\text{ or }\underset{\mathrm{h}\to 0^{+}}{lim}\mathrm{f}(\mathrm{a}+\mathrm{h})=\mathrm{f}(\mathrm{a})\text{ or }RHL=f(a)$
For example,
$\mathrm{f}(\mathrm{x})=\mathrm{y}=[\mathrm{x}],($ where, [.] is G.I.F $)$. is right continuous at $\mathrm{x}=2$

$f(2)=\underset{x\to 2^{+}}{lim}[x]=2$
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 $c\in (a,b),f(x)$ is continuous if

$\underset{x\to c^{}}{lim}f(x)=\underset{x\to c^{+}}{lim}f(x)=f(c)$
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 rightcontinuous at $x=a$ and
is leftcontinuous at $x=b$.
i.e.At $\mathrm{x}=\mathrm{a}$, we need to check $f(a)=\underset{x\to a^{+}}{lim}f(x)(=\underset{h\to 0^{+}}{lim}f(a+h)=$ 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 $f(b)=\underset{x\to b^{}}{lim}f(x)(=\underset{h\to 0^{+}}{lim}f(bh)=$ 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 $\mathrm{x}=\mathrm{c}:\underset{x\to {\mathrm{c}}^{}}{lim}[x]=2$ (as in close left neighbourhood of $\mathrm{x}=\mathrm{c}$, the function equals 2 )
RHL at $\mathrm{x}=\mathrm{c}$ : $\underset{x\to c^{+}}{lim}[x]=2$ (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 $x=2$
$f(2)=2$

$\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}[x]=\underset{h\to 0^{+}}{lim}[2+h]=2$
So $f(x)$ is left continuous at $x=2$

Condition 3

Left continuity at $x=3$
$f(3)=3$ and

$\underset{x\to 3^{}}{lim}[x]=\underset{h\to 0^{+}}{lim}[3h]=2$

(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 $\underset{x\to a}{lim}f(x)$ 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. $\underset{x\to a}{lim}f(x)=f(a)$.

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, $x\ne 2$

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

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

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