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3. LIMIT AND CONTINUITY
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3.1 Limit Function at One Point
An intuitive introduction
Let
11)(
2
= xxxf
This function is not defined at x=1 (0/0), it still makes sense to ask
what happens to values of f(x) as x moves along x-axis toward 0.
The following table shows a succession of values of f(x) correspondingto succession of values of x that move toward 0
x
f(x)
0.9 0.99 0.999 1.11.011.0010.9999 1.00011
21.9 1.99 1.999 1.9999 2.0001 2.001 2.01 2.1
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1
2
x x
f(x)
f(x)
geometrically
The result in the table suggest thatthe values of f(x) approach 1
We denote this by writing
21
1lim
2
1 =
x
x
x
we read : the limit of as x approaches 1 is equal to 2
1
12
x
x
Definition (intuitive) . It mean, the value of f(x) approaches thenumber Las x approaches c
Lxfcx
=
)(lim
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Lxfcx
=
)(lim
L
c
L
L
+L
there is such that0>
c
L
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)(lim xfcx +
The Limit from the right and left(one-sided limit)
cx
Ifxapproaches cfrom the left side (fromnumber on which is smaller than c), weget the left-hand limit, denoted by
)(lim xfcx
c x
lim ( ) lim ( ) lim ( )andx c x c x c
f x L f x L f x L +
= = =
The relationship among two-side limit an one-side limit
If )(lim xfcx
+)(lim xf
cx
then does not exist)(lim xfcx
If x approaches c from the right side (fromnumber on which is greater than c), we
get the right-hand limit, denoted by
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+
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)(lim0
xfx
0lim 20 == xx
)(lim0
xfx +
0lim0
==+x
x
0)(lim0
=
xfx
)(lim1
xfx
1lim
1==
x
x
)(lim1 xfx + 32lim2
1 =+= + xx
)(lim1
xfx
)(lim2 xfx 62lim
2
2=+=
xx
does not exist
c. The formula of f(x) does not change at, thus
b. Because the formula of f(x) change at x = 1 then first findingthe one-sided limit at point x = 1.
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d.
For x 02)( xxf =
Graph: parabola
For 0
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2. Find the value of c so that
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Problems
>+
+=
1,2
1,1)(
2
2
xxx
xxxf
)(lim1
xfx
x
f x
+1lim ( )
x
f x
1
lim ( )
xxxg 32)( =
x
g x
2lim ( )
x
g x
+2
lim ( )x
g x2
lim ( )
22)(
=
xxxf
x
f x
2
lim ( )x
f x
+2
lim ( )x
f x2
lim ( )
1. Let
a.Evaluate and
b. Does exist? if it limit exist evaluate
2. Let , evaluate
3. Let , evaluate
a. b. c.
a. b. c.
B.
x
f x
1
lim ( )
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lim ( ) lim ( )and x a x a
f x L g x G
= =
[ ] GLxgxfxgxfaxaxax
==
)(lim)(lim)()(lim
TheoremLet
then
[ ] LGxgxfxgxfaxaxax
==
)(lim)(lim)()(lim
0,)(lim
)(lim
)(
)(lim ==
Gbila
G
L
xg
xf
xg
xf
ax
ax
ax
2.
3.
4.
n
ax
n
ax xfxf ))(lim())((lim =
,n positive integer
nn
ax
n
axLxfxf ==
)(lim)(lim5. provided if n is even
1.
0L
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222 )1(1
1sin)1()1(
x
xxx
)()()( xhxgxf
lim ( ) , lim ( ) x c x c
f x L h x L
= =
Lxgcx
=
)(lim
1
1
sin)1(lim
2
1
xxx
The Pinching theorem
Let for all x in some open interval containing the point c and
then
Example Evaluate
Because 1)1
1sin(1
x
and 0)1(lim 21
=
xx
0)1(lim, 21
=
xx
0
1
1sin)1(lim 2
1=
x
xx
then
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Limits of Trigonometric Functions
1sin
lim.10
= x
x
x
1coslim.2 0=
xx
1tan
lim.30
= x
x
x
Example
2.
2
2tan5
4.4
4sin3
lim2tan5
4sin3lim
00
x
xx
x
xx
xx
xx
+=
+
2.
2
2tanlim5
4.4
4sinlim3
0
0
x
xx
x
x
x
+=
3
7
2.2
2tanlim5
4.4
4sinlim3
02
04=
+=
x
xx
x
x
x
x 0 if only if 4x 0
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Problems
tt
t sin1coslim
2
0 +
t
tt
t sec2
sincotlim
0
t
t
t 2
3tanlim
2
0
tt
tt
t sec
43sinlim
0
+
Evaluate
1.
2.
3.
4.
x
x
x 2sin
tanlim
05.
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Example Evaluate
1
1lim
2
1
+ x
x
x
a.1
1lim
2
2
1
+ x
x
x x
x
x sinlim
+b. c.
Solution
a. 021lim 21
>=+x
x,g(x)=x-1 approaches 0 from the downward
thus
=
+ 1
1
lim
2
1 x
x
x
b. 021lim2
1>=+
x
x
approaches 0 from the upward1)( 2 = xxg
thus
+=
+ 1
1lim
2
2
1 x
x
x
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c.
0lim >=+
xx
and
f(x)=sinx
x
If x approaches from the right then sin(x) approaches 0 fromdownward
=+ xx
x sinlim
thus
Because
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Limits at Infinity
Lxfx
=
)(lima. if >> |)(|00 LxfMxM
or f(x) approaches L if x increase without boundL
x
Example Evaluate
42
52lim
2
2
+
++
x
xx
x
Solution
)2(
)1(lim
2
2
42
522
x
xx
x x
x
+
++=
42
52lim
2
2
+
++ x
xx
x
2
2
42
521
lim
x
xxx
+
++=
= 1/2
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Lxfx
=
)(lim if
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Example Evaluate
xxxx
+++
3lim 2
Solution:
If x , It is form ( )
xxxx
+++
3lim 2 )
3
3(3lim
2
22
xxx
xxxxxx
x
++
+++++=
xxx
xxx
x ++
++=
3
3lim
2
22
xxx
x
x ++
+=
3
3lim
2
xx
x
xx
x
x ++
+
= )1(
)1(
lim2
312
3
||2 xx =
xxx
xx
x
x +++=
2
31
3
1)1(lim
2
1
)11(
1lim
2
31
3
=+++
+=
xx
x
x
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Problems
limx
x
x +
+
3
3
3
limx x + 2
2
3
4
)1(lim xxx
limx
x
x +1 2
1
1lim
2
+ x
x
x
limx
x x
x
+
+
2
1
.
Evaluate
1.
2.
3.
4.
5.
6.
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Continuity
A Function f(x) is said continuity at x= a if
(i) f(a) is defined
lim ( )x a f x exist(ii)
(iii) )()(lim afxfax
=
If one or more of the conditions fails to hold, then f is called discontinuous
at a and ais called apoint of discontinuity off
a
(i)
f(a) does not exist
f is not continuous at
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a
(ii)
1L
2LLeft-hand limit right-hand limit ,Thus limit does not exist at x=a
f is not continuous at x=a
(iii)
a
f(a)f(a) is defined
)(lim xfax
L exist
f is not continuous at x=a
lim ( ) ( )x a
f x f a
but
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(iv)
a
f(a)
f(a) is defined)(lim xf
axexist
)()(lim afxfax
=
f(x) is continuous x=a
Removable discontinuity
For case (i) discontinuity can be removedby define f(a) = limit of function at a
a
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ExampleDetermine whether f is continuous at x=2
2
4)(
2
=
x
xxf
=
=
2,3
2,
2
4)(
2
x
x
x
xxfa. b.
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Solution :f(x) is continuous at x=2, if
f is continuous from the left at x=2
)2()(lim2 fxfx=
aaxxfxx
+=+=
2lim)(lim22
1412)2( 2 == aaf
2 + a = 4a 1-3a = -3
a = 1
f is continuous from the right at x=2
)2()(lim2
fxfx
=+1412)2( 2 == aaf
141lim)(lim 222
==++
aaxxfxx
trivial
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1. Let
>+
=
1,22
1,1)(
2
xx
xxxf
Determine whether f is continuous at x= -1
Problems
2. Find the value of a + 2b so that f(x) will be continuous everywhere
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Continuity on a Closed Interval
A functionf(x) is said to be continuous on closed interval [a,b] if
1.f(x) is continuous on ( a,b )
2.f(x) is continuous from the right atx = a
3.f(x) is continuous from the left atx = b
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Theorem Polynomials are continuous function
A rational function is continuous everywhere except at the points
where the denominator is zero
Let , then
f(x) is continuous everywhere ifn is odd
f(x) is continuous for positive number if n is even
Example where is continuous ?
From theorem, f(x) is continuous for x-4> 0 or x > 4.
f(x) is continuous from the
right at x=4
Thus f(x) is continuous at [4, )
n xxf =)(
4)( = xxf
)4(04lim)(lim44
fxxfxx
===++
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f xx x
x( ) =
+
+
2 3
3
f xx
x( ) =
2
3
4
8
f xx
x( )
| |=
2
2
94
1)(
2
=
x
xxf
24)( xxxf =
A. Find the points of discontinuity, if any
B. where is f(x) continuous
Problems
1.
2.
3.
1.
2.
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Limit and continuity of Composition Function
Theorema
If andf(x) is continuous atL, then
Theorema
If g(x) is continuous at a,f(x) is continuous at g(a), then the composition
is continuous at a
Proof
( f is continuous at g(a) )= f(g(a)) ( g is continuous at a )
= (fog)(a)
Lxgax
=
)(lim
)())(lim())((lim Lfxgfxgfaxax
==
))(( xgf
))((lim))((lim xgfxgfaxax
=
))(lim( xgf ax=
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+
+=
43
13
cos)( 2
4
xx
xx
xf
))(()( xhgxf =
43
13
)( 2
4
+
+
= xx
xx
xh and g(x) = cosx
Example Where is continuous?
Solution:
f(x) can be written as
where
Because h(x) is continuous on R-{-4,1} and g(x) is continuous everywherethen f(x) is continuous on R-{-4,1}
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