Continuity of a Function

when we say that the function f(x) is continuous at a point x=a, it means that at point (a, f(a)) the graph of the function has no holes or gaps. That is,its graph is unbroken at a point (a,f(a) )

A function f(x) is said to be continuous at x=a

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

(i.e.), L.H.L.= R.H.L. = value of a function at x=a

or $\displaystyle \lim _{ x\longrightarrow a }{ f\left( x \right) } =f\left( a \right)$

If f(x) is not continuous at x=a we say that f(x) is discontinuous at x=a in any of the following cases:

$\displaystyle \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } and\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) }$ exists but are not equal
$\displaystyle \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } and\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) }$ exists and are equal but not equal to f(a)
f(a) is not defined
At least one of the limit does not exists

Graphical View

$\displaystyle \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } and\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) }$ exists but are not equal $\displaystyle Here,\lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } ={ l }_{ 1 }\\ \lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } ={ l }_{ 2 }\\ \therefore \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } \quad and\quad \lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) }$ exists but are not equal.Thus f(x) discontinuous at x=a It does not matter whether f(a) exists or not.

Example

$\displaystyle If\quad f\left( x \right)=\frac { \left| x \right| }{ x }$ .Discuss the continuity at $\displaystyle x\longrightarrow 0$

$\displaystyle Here\quad f\left( x \right) =\frac { \left| x \right| }{ x } \\ \therefore R.H.L.\quad at\quad x\longrightarrow 0\\ Let\quad x=0+h\\ i.e.\quad \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\longrightarrow { 0 } }{ f\left( 0+h \right) } =\lim _{ x\longrightarrow { 0 } }{ \frac { \left| 0+h \right| }{ 0+h } } =\lim _{ x\longrightarrow { 0 } }{ \frac { h }{ h } } =1\\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =1\\ again\quad L.H.L.\quad at\quad x\longrightarrow 0\\ let\quad x=0-h\\ i.e.\quad \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ x\longrightarrow { 0 } }{ f\left( 0-h \right) } =\lim _{ x\longrightarrow { 0 } }{ \frac { \left| 0-h \right| }{ 0-h } } =\lim _{ x\longrightarrow { 0 } }{ \frac { h }{ -h } } =-1\quad \\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =-1\\ \therefore \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } \neq \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } .\quad Thus\quad f\left( x \right)\quad is\quad discontinous\quad at\quad x\longrightarrow 0$

Graphically
$\displaystyle \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } \quad and\quad \lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) }$ exists and are equal but equal to f(a) $\displaystyle Here\lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } =\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } =L$ f(a) is also defined but $\displaystyle f\left( a \right)\neq L$ So again,limit of f(x) exists at x=aBut it is not continuous at x=a

Example

$\displaystyle If\quad f\left( x \right) =2x+3\quad when\quad x<0\\ =0\quad when\quad x=0\\ ={ x }^{ 2 }+3\quad when\quad x>0$ Discuss the continuity

Solution:

$\displaystyle If\quad f\left( x \right) =2x+3\quad when\quad x<0\\ =0\quad when\quad x=0\\ ={ x }^{ 2 }+3\quad when\quad x>0\\ \therefore R.H.L.\quad at\quad x=0,\quad \\ Let\quad x=0+h\\ i.e.\quad \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ h\longrightarrow { 0 } }{ f\left( 0+h \right) } =\lim _{ h\longrightarrow { 0 } }{ \left\{ { \left( 0+h \right) }^{ 2 }+3 \right\} } =3\\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =3\\ again,\quad L.H.L.\quad at\quad x=0\\ Let\quad x=0-h\\ i.e.\quad \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ h\longrightarrow { { 0 }^{ + } } }{ f\left( 0-h \right) } =\lim _{ h\longrightarrow { 0 } }{ \left\{ 2{ \left( 0+h \right) }+3 \right\} } =3\\ \Rightarrow \lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } =3\\ But\quad f\left( 0 \right) =0\\ \therefore \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\longrightarrow { 0 }^{ - } }{ f\left( x \right) } \neq f\left( 0 \right)$
f(a) is not defined
Here, $\displaystyle \lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } =L\quad and\quad \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } =L$
But f(a) is not defined.So, f(x) is discontinuous at x=a

Example

if $\displaystyle f\left( x \right) =\frac { { x }^{ 2 }-1 }{ x-1 }$ Discuss the continuity at $latexdisplaystylex\longrightarrow 1 &s=1 $
Solution:

$\displaystyle f\left( x \right) =\frac { { x }^{ 2 }-1 }{ x-1 } \\ \lim _{ x\longrightarrow 1 }{ f\left( x \right) } =\quad \lim _{ x\longrightarrow 1 }{ \frac { { x }^{ 2 }-1 }{ x-1 } } \\ =\frac { \left( x-1 \right) \left( x+1 \right) }{ \left( x-1 \right) } \\ =\lim _{ x\longrightarrow 1 }{ \left( x+1 \right) } \\ =2\\ But\quad f\left( 1 \right)=\frac { 0 }{ 0 } \quad \left( Indetermined\quad form \right) \\ \therefore f\left( 1 \right)\quad is\quad not\quad defined\quad at\quad x=1\\ Hence\quad f\left( x \right)\quad is\quad discontinuous\quad at\quad x=1$
Graphically:
Which shows $\displaystyle \lim _{ x\longrightarrow 1 }{ f\left( x \right) } =2$ but f(1) is not defined.
So f(x) is discontinuous at x=1