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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.

(B)Discontinuity of 2nd kind:

If at least one of \displaystyle \lim _{ h\longrightarrow 0 }{ f\left( a+h \right) } and\quad \lim _{ h\longrightarrow 0 }{ f\left( a-h \right) }

is non- existent or infinite then f(x) is said to have discontinuity and 2nd kind at x=a

Example : Show the function,

\displaystyle f\left( x \right)=\frac { 1 }{ \left| x \right| }    has discontinuity at 2nd kind at x=0

Solution: We have

\displaystyle f\left( x \right) =\frac { 1 }{ \left| x \right| } =\infty 

Which shows function has discontinuity of 2nd kind.

Graphically:

Here the graph is broken at x=0 as

\displaystyle x\longrightarrow 0\Rightarrow f\left( x \right) \longrightarrow \infty

Here the graph is broken at x=0 as \displaystyle x\longrightarrow 0\Rightarrow f\left( x \right) \longrightarrow \infty

Therefore f(x) is discontinuity of 2nd kind.

April 18, 2019
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