Kinds of Discontinuity

We are now going to look at the two main types of discontinuities that can arise in a function. You should be able to distinguish between each type of discontinuity when a function $f$ may contain each type of discontinuity.

Discontinuity is of two kinds listed as,

(A) Discontinuity of 1st kind:

(i) First kind removable discontinuity

(ii) Non-removable discontinuity or jump discontinuity

(i) First kind removable discontinuity

If $\displaystyle \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } =\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } \neq f\left( a \right)$

then f(x) is said to have first kind of removable discontinuity:

Example : Examine the function

$\displaystyle f\left( x \right) =x-1,\quad when\quad x<0\\= \frac { 1 }{ 4 } ,\quad when\quad x=0\\= { x }^{ 2 }-1,\quad when\quad x>1$

Solution:

Graphically,f(x) could be plotted as ,

Which shows $\displaystyle \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =0,butf\left( 0 \right)=\frac { 1 }{ 4 }$

Thus ,f(x) could be made continuously taking.
$\displaystyle f\left( x \right) =x-1,\quad x\textless 0\\ =-1,\quad x=0\\ ={ x }^{ 2 }-1,\quad x\textgreater 0$
so, we could say f(x) becomes continuous, if
$\displaystyle f\left( x \right) =x-1,\quad x\textless 0\\ =-1,\quad x=0\\ ={ x }^{ 2 }-1,\quad x\textgreater 0$

(ii) Non-Removable discontinuity:

$\displaystyle If\quad \lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } \neq \lim _{ x\longrightarrow a }{ f\left( x \right) }$
Then f(x) is said to have First kind non-removable discontinuity.

Example : Show the function,

$\displaystyle f\left( x \right) =\frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } ,\quad when\quad x\neq 0\\ =0,\quad when\quad x=0$
has non-removable discontinuity at x=0

Solution: We have

$\displaystyle f\left( x \right) =\frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } ,\quad when\quad x\neq 0\\ =0,\quad when\quad x=0$

then R.H.L. at x=0, Let x=0+h

$\displaystyle \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ h\longrightarrow 0 }{ f\left( 0+h \right) } =\lim _{ h\longrightarrow 0 }{ \frac { { e }^{ \frac { 1 }{ 0+h } }-1 }{ { e }^{ \frac { 1 }{ 0+h } }+1 } } =\lim _{ h\longrightarrow 0 }{ \frac { { e }^{ \frac { 1 }{ h } }-1 }{ { e }^{ \frac { 1 }{ h } }+1 } } \\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ h\longrightarrow 0 }{ \frac { 1-\frac { 1 }{ { e }^{ \frac { 1 }{ h } } } }{ 1+\frac { 1 }{ { e }^{ \frac { 1 }{ h } } } } } \\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } \frac { 1-0 }{ 1+0 } =1\qquad \left[ as\quad h\longrightarrow 0;\frac { 1 }{ h } \longrightarrow \infty \Rightarrow { e }^{ \frac { 1 }{ h } }\longrightarrow \infty ;\frac { 1 }{ { e }^{ \frac { 1 }{ h } } } \longrightarrow 0 \right] \\ \therefore \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =1$

again, L.H.L. at x=0 , at x=0, Let x=0-h

$\displaystyle \Rightarrow \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ h\longrightarrow 0 }{ f\left( 0-h \right) } =\lim _{ h\longrightarrow 0 }{ \frac { { e }^{ -\frac { 1 }{ h } }-1 }{ { e }^{ -\frac { 1 }{ h } }+1 } } =\frac { 0-1 }{ 0+1 } =-1\\ \therefore \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =-1$ \qquad \left[ as\quad h\longrightarrow 0;{ e }^{ -\frac { 1 }{ h } }\longrightarrow 0 \right] \\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } \neq \lim _{ x\longrightarrow { 0 }^{ – } }{ f\left( x \right) } &s=1 $
Thus,f(x) has non-removable discontinuity.