Continuity

Md. Masum BillahAssistant Professor of Mathematics

Department of Arts and SciencesAUST

1

Continuity of a Function

A function f is continuous at the point x = a if the followings are true:

) ( ) is definedi f a

) lim ( ) existsx a

ii f x

) lim ( ) ( )x a

iii f x f a

a

f(a)

y

x

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ExampleIs the function f given by f (x) x2 5

continuous at x = 3? Why or why not?

f (3) 32 5 9 5 41)

2) By the Theorem on Limits of Rational Functions,

limx3

x2 5 32 5 9 5 4

3) Since f is continuous at x = 3.

limx3

f (x) f (3)

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Physical phenomena are usually continuous.

Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it.

The graph can be drawn without removing your pen from the paper.

For instance, the displacement or velocity of a vehicle varies continuously with time.

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Properties of Continuous Functions

The constant function f (x) is continuous everywhere. Ex. f (x) = 10 is continuous everywhere.

The identity function f (x) = x is continuous everywhere.

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Properties of Continuous Functions

A polynomial function y = P(x) is continuous at everywhere.

A rational function is continuous

at all x values in its domain.

( )( )( )

p xR xq x

If f and g are continuous at x = a, then

, , and ( ) 0 are continuous

at .

ff g fg g ag

x a

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xxf sin)( is continuous for every x.Show that

xxf sin)( and we take an arbitrary value of x say x=a. )(lim xf

axexists or not.Now, we have to check

)sin(lim0

hah

sinh)coscosh(sinlim0

aah

0.cos1.sin aa asin

)(lim xfax

)(lim0

hafh

L.H.L. =

Problem

Solution

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)sin(lim0

hah

sinh)coscosh(sinlim0

aah

0.cos1.sin aa asin

)(lim xfax

)(lim0

hafh

R.H.L. =

)(..........sin)(lim iaxfax

afxf

iiaafxxf

ax

limthat(ii)and(i)fromthatobserveWe

).(..........sinsin)(Now

asinSince L.H.L. = R.H.L.

Since f(x) satisfies all conditions of continuity, thus the function f(x) is continuous at 8M. M. Billah

A function f(x) is defined as follows:

f(x) = cosx , if x 0- cosx , if x < 0

Is f(x) is continuous at x = 0 ?

)(lim0

xfx

)0(lim0

hfh

)(lim0

hfh

)cos(lim0

hh

1

coshlim0

h

When x

Since L.H.L R.H.L

Thus, the function f(x) is not continuous at x = 0.

)(lim0

xfx

So, does not exist.

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1211

210

)(xwhenx

xwhenxxf

21x

Examine the continuity of the following function at x =

When then f(x) = x

)(lim21

xfx

)21(lim

0hf

h

)21(lim

0h

h

210

21

L.H.L=

21x

)(lim21

xfx

)

21(lim

0hf

h

h

h 211lim

0

h

h 21lim

0

Again, when then f(x) = 1 x

R.H.L=

21

Problem

Solution

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21)(lim

21

xfx

21

211

21

f

21)(limSince,Again

21

fxfx

Since L.H.L = R.H.L.

21xNow, when then f(x) = 1 x

21

So f(x) is continuous at x =

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4,167

4,32)( x

x

xxxf

)(lim4

xfx

Determine whether the following function is continuous at x=4

4x )167()( xxf thenWhen

Now R.H.L

)4(lim0

hfh

hh 4167lim

0

= 7+4 = 11

4x 32)( xxfthenWhen

)(lim4

xfx

Now L.H.L

)4(lim0

hfh

3)4(2lim0

hh

38 11

11)(lim4

xfx

Since L.H.L = R.H.L.

Problem

Solution

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11

3424

f

4)(limSince,Again4

fxfx

4xNow, when then f(x) = 2x+3

So f(x) is continuous at x = 4

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Compute the value for the constant of k, that will make the following function continuous at x=1

1,1,27

)( 2 xkxxx

xf

)(lim1

xfx

1x 2)( kxxf thenWhen

Now R.H.L

)1(lim0

hfh

1x 27)( xxfthenWhen

)(lim1

xfx

Now L.H.L

)1(lim0

hfh

2

0)1(lim hk

h

k

2)1(7{lim0

hh

5

Problem

Solution

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1x 27)( xxfthenWhen

5217)1( f

5)(lim)(lim11

xfxfxx

If the function is continuous than

So, k = 5.

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Discuss the continuity of the function

21;310;02;

)(xxxxxx

xf

At x = 0 and x = 1.

Problem

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Now, lets see how to detect discontinuities when a function is defined by a formula.

Where are each of the following functions discontinuous?

2 2( )2

x xf xx

2

1 0( )

1 0

if xf x x

if x

2 2 2( ) 21 2

x x if xf x xif x

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2 2( )2

x xf xx

Notice that f(2) is not defined.So, f is discontinuous at 2.

The kind of discontinuity illustrated here is called removable.

We could remove the discontinuity by redefining fat just the single number 2.

The function is continuous.( ) 1g x x

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21 0( )

1 0

if xf x x

if x

Here, f(0) = 1 is defined.

However,

20 0

1lim ( ) limx x

f xx

does not exist.

So, f is discontinuous at 0.The discontinuity is called an infinite discontinuity.

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22 2

2

2

2lim ( ) lim2

( 2)( 1)lim2

lim( 1) 3

x x

x

x

x xf xx

x xx

x

2lim ( ) (2)x

f x f

So, f is not continuous at 2.

exists.

Here, f(2) = 1 is defined and

2 2 2( ) 21 2

x x if xf x xif x

The kind of discontinuity illustrated here is called removable.

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The function jumps from one value to another.

These discontinuities are called jump discontinuities.

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From the appearance of the graphs of the sine and cosine functions, we would certainly guess that they are continuous.

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is continuous except where cos x = 0.sintancos

xxx

This happens when x is an odd integer multiple of .2

So, y = tan x has infinite discontinuities when

3 52, 2, 2,x

and so on.

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DifferentiabilityA function f(x) is said to be differentiable at a point x = a if

h

afhafafh

)()(lim0

exist.

h

afhafafh

)()(lim0

Again will exist if

hafhaf

h

)()(lim0

h

afhafh

)()(lim0

and

].af [L Derivative HandLeft called is )()(lim0

h

afhafh

In this case,

].af [R Derivative HandRight called is )()(lim0

h

afhafh

and

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Show that f(x) is continuous at x=0, but not differentiable at x=0.

23

23

0,230,23

)(xxxx

xf

0x xxf 23)( thenWhen

Now R.H.D

0x xxf 23)( thenWhen

)0(fL Now L.H.D

hfhf

h

)0()0(lim0

hfhf

h

)0()(lim0

hh

h

)0.23()23(lim0

22lim0

h

hh

)0(fR

hfhf

h

)0()0(lim0

hfhf

h

)0()(lim0

hh

h

)0.23()23(lim0

22lim0

h

hh

Problem

Solution

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00Since fRfL The function f(x) is not differentiable at x = 0.

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0if,0

0if,1sin)(2

x

xx

xxf

Show that f(x) is differentiable at x = 0 but f(x) is not continuous at x = 0

Now R.H.D)0(fL Now L.H.D

hfhf

h

)0()0(lim0

)0(fR

hfhf

h

)0()0(lim0

hh

hh

01sinlim

2

0

hh

h

1sinlim0

1to1betweenvaluany0 0

hh

h

h

01sinlim

2

0

hh

h

1sinlim0

1to1betweenvaluany0 0

Problem

Solution

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00Since fRfL The function f(x) is differentiable at x = 0.

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