Download - Continuity bY MM BILLAH
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Continuity
Md. Masum BillahAssistant Professor of Mathematics
Department of Arts and SciencesAUST
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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
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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
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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.
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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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