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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
2M. M. Billah
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)
3M. M. Billah
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.
4M. M. Billah
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.
5M. M. Billah
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
6M. M. Billah
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
7M. M. Billah
)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.
10M. M. Billah
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
11M. M. Billah
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 =
12M. M. Billah
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
13M. M. Billah
11
3424
f
4)(limSince,Again4
fxfx
4xNow, when then f(x) = 2x+3
So f(x) is continuous at x = 4
14M. M. Billah
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
15M. M. Billah
1x 27)( xxfthenWhen
5217)1( f
5)(lim)(lim11
xfxfxx
If the function is continuous than
So, k = 5.
16M. M. Billah
Discuss the continuity of the function
21;310;02;
)(xxxxxx
xf
At x = 0 and x = 1.
Problem
17M. M. Billah
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
18M. M. Billah
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
19M. M. Billah
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.
20M. M. Billah
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.
21M. M. Billah
The function jumps from one value to another.
These discontinuities are called jump discontinuities.
22M. M. Billah
From the appearance of the graphs of the sine and cosine functions, we would certainly guess that they are continuous.
23M. M. Billah
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.
24M. M. Billah
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
25M. M. Billah
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
26M. M. Billah
00Since fRfL The function f(x) is not differentiable at x = 0.
27M. M. Billah
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
28M. M. Billah
00Since fRfL The function f(x) is differentiable at x = 0.
29M. M. Billah