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Universitas Negeri Jakarta
Kalkulus I
Drs. Tasman Abbas
Sesion #01-14
Jurusan FisikaFakultas Matematika dan Ilmu Pengetahuan Alam
UNIVERSITAS NEGERI JAKARTA
Outline• Definition• Evaluation of Limits• Continuity• Limits Involving Infinity
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Limits and Continuity
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Limit
We say that the limit of ( ) as approaches is and writef x x a L
lim ( ) x a
f x L
if the values of ( ) approach as approaches . f x L x a
a
L( )y f x
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Limits, Graphs, and Calculators 21
11. a) Use table of values to guess the value of lim1x
xx
2
1b) Use your calculator to draw the graph ( ) 1
xf xx
and confirm your guess in a)2. Find the following limits
0
sina) lim by considering the values x
xx
1, 0.5, 0.1, 0.05, 0.001. x Thus the limit is 1. sinConfirm this by ploting the graph of ( ) xf xx
5
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0
b) limsin by considering the values x x
1 1 1(i) 1, , , 10 100 1000x
2 2 2(ii) 1, , , 3 103 1003x
This shows the limit does not exist.
Confrim this by ploting the graph of ( ) sin f x x
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c) Find2
3 if 2lim ( ) where ( )
1 if 2x
x xf x f x
x
-2
62 2
lim ( ) = lim 3x x
f x x
Note: f (-2) = 1
is not involved
23 lim
3( 2) 6x
x
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2
2
4( 4)a. lim2x
xx
0
1, if 0b. lim ( ), where ( )
1, if 0x
xg x g x
x
20
1c. lim ( ), where f ( )xf x x
x
0
1 1d. limx
xx
Answer : 16
Answer : no limit
Answer : no limit
Answer : 1/2
3) Use your calculator to evaluate the limits
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The Definition of Limit-
lim ( ) We say if and only if x a
f x L
given a positive number , there exists a positive such that
if 0 | | , then | ( ) | . x a f x L
( )y f x a
LL
L
a a
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such that for all in ( , ), x a a a
then we can find a (small) interval ( , )a a
( ) is in ( , ).f x L L
This means that if we are given a
small interval ( , ) centered at , L L L
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Examples2
1. Show that lim(3 4) 10.x
x
Let 0 be given. We need to find a 0 such that if | - 2 | ,x then | (3 4) 10 | .x
But | (3 4) 10 | | 3 6 | 3 | 2 |x x x
if | 2 |3
x So we choose .
3
1
12. Show that lim 1.x x
Let 0 be given. We need to find a 0 such that 1if | 1| , then | 1| .x x
1 11But | 1| | | | 1| . x xx x x
What do we do with the x?
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1 31If we decide | 1| , then . 2 22x x
1And so <2. x
1/2
11Thus | 1| | 1| 2 | 1| . x xx x
1 Now we choose min , .3 2
1 3/2
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The right-hand limit of f (x), as x approaches a, equals L
written:
if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the right of a.
lim ( )x a
f x L
a
L( )y f x
One-Sided Limits
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The left-hand limit of f (x), as x approaches a, equals M
written:
if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the left of a.
lim ( )x a
f x M
a
M
( )y f x
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2 if 3( )2 if 3x xf xx x
1. Given
3lim ( )x
f x
3 3lim ( ) lim 2 6x x
f x x
2
3 3lim ( ) lim 9x x
f x x
Find
Find 3
lim ( )x
f x
Examples of One-Sided Limit
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1, if 02. Let ( )
1, if 0.x x
f xx x
Find the limits:
0lim ( 1)x
x
0 1 1 0
a) lim ( )x
f x
0b) lim ( )
xf x
0lim( 1)x
x
0 1 1
1c) lim ( )
xf x
1lim( 1)x
x
1 1 2
1d) lim ( )
xf x
1lim( 1)x
x
1 1 2
More Examples
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lim ( ) if and only if lim ( ) and lim ( ) .x a x a x a
f x L f x L f x L
For the function
1 1 1lim ( ) 2 because lim ( ) 2 and lim ( ) 2.x x xf x f x f x
But
0 0 0lim ( ) does not exist because lim ( ) 1 and lim ( ) 1.x x xf x f x f x
1, if 0( )
1, if 0.x x
f xx x
This theorem is used to show a limit does not exist.
A Theorem
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Limit TheoremsIf is any number, lim ( ) and lim ( ) , then
x a x ac f x L g x M
a) lim ( ) ( )x a
f x g x L M
b) lim ( ) ( ) x a
f x g x L M
c) lim ( ) ( )x a
f x g x L M
( )d) lim , ( 0)( )x a
f x L Mg x M
e) lim ( )x a
c f x c L
f) lim ( ) n n
x af x L
g) lim x ac c
h) lim
x ax a
i) lim n n
x ax a
j) lim ( ) , ( 0)
x af x L L
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Examples Using Limit RuleEx. 2
3lim 1x
x
2
3 3lim lim1x xx
2
3 3
2
lim lim1
3 1 10x xx
Ex.1
2 1lim3 5x
xx
1
1
lim 2 1
lim 3 5x
x
x
x
1 1
1 1
2lim lim1
3lim lim5x x
x x
x
x
2 1 13 5 8
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More Examples
3 31. Suppose lim ( ) 4 and lim ( ) 2. Find
x xf x g x
3
a) lim ( ) ( ) x
f x g x
3 3 lim ( ) lim ( )x x
f x g x
4 ( 2) 2
3
b) lim ( ) ( ) x
f x g x
3 3 lim ( ) lim ( )x x
f x g x
4 ( 2) 6
3
2 ( ) ( )c) lim ( ) ( )x
f x g xf x g x
3 3
3 3
lim 2 ( ) lim ( )
lim ( ) lim ( )x x
x x
f x g x
f x g x
2 4 ( 2) 5
4 ( 2) 4
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Indeterminate forms occur when substitution in the limit results in 0/0. In such cases either factor or rationalize the expressions.
Ex. 25
5lim25x
xx
Notice form00
5
5lim5 5x
xx x
Factor and cancel common factors
5
1 1lim5 10x x
Indeterminate Forms
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9
3a) lim 9x
xx
9
( 3)( 3)
( 3) = lim( 9)x
xx
xx
9
9 lim( 9)( 3)x
xx x
9
1 1 lim63x x
2
2 3 2
4b) lim 2x
xx x
2 2
(2 )(2 )= lim(2 )x
x xx x
2 2
2 = limx
xx
2
2 ( 2) 4 1( 2) 4
More Examples
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The Squeezing TheoremIf ( ) ( ) ( ) when is near , and if f x g x h x x a
, then lim ( ) lim ( )x a x a
f x h x L
lim ( ) x ag x L
2
0 Show that liExampl m 0e: .
xx sin x
0
Note that we cannot use product rule because limxsin x
DNE!
But 1 sin 1 x 2 2 2and so sin . x x xx
2 2
0 0Since lim lim( ) 0,
x xx x
we use the Squeezing Theorem to conclude
2
0lim 0.xx sin x
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Continuity A function f is continuous at the point x = a if the following are true:
) ( ) is definedi f a) lim ( ) existsx a
ii f x
a
f(a)
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A function f is continuous at the point x = a if the following are true:
) ( ) is definedi f a) lim ( ) existsx a
ii f x
) lim ( ) ( )x a
iii f x f a
a
f(a)
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At which value(s) of x is the given function discontinuous?1. ( ) 2f x x
2
92. ( )3
xg xx
Continuous everywhere
Continuous everywhere except at 3x
( 3) is undefinedg
lim( 2) 2 x a
x a
and so lim ( ) ( )x a
f x f a
-4 -2 2 4
-2
2
4
6
-6 -4 -2 2 4
-10
-8
-6
-4
-2
2
4
Examples
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2, if 13. ( )
1, if 1x x
h xx
1lim ( )xh x
and
Thus h is not cont. at x=1.
11
lim ( )xh x
3
h is continuous everywhere else
1, if 04. ( )
1, if 0x
F xx
0lim ( )x
F x
1 and
0lim ( )x
F x
1
Thus F is not cont. at 0.x F is continuous everywhere else
-2 2 4
-3
-2
-1
1
2
3
4
5
-10 -5 5 10
-3
-2
-1
1
2
3
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Continuous Functions
A polynomial function y = P(x) is continuous at every point x.
A rational function is continuous at every point x in its domain.
( )( ) ( )p xR x q x
If f and g are continuous at x = a, then
, , and ( ) 0 are continuous
at
ff g fg g agx a
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Intermediate Value TheoremIf f is a continuous function on a closed interval [a, b] and L is any number between f (a) and f (b), then there is at least one number c in [a, b] such that f(c) = L.
( )y f x
a b
f (a)
f (b)L
c
f (c) =
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Example
2Given ( ) 3 2 5,Show that ( ) 0 has a solution on 1,2 .
f x x xf x
(1) 4 0(2) 3 0ff
f (x) is continuous (polynomial) and since f (1) < 0 and f (2) > 0, by the Intermediate Value Theorem there exists a c on [1, 2] such that f (c) = 0.
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Limits at Infinity
For all n > 0,1 1lim lim 0n nx xx x
provided that is defined.1nx
Ex.2
23 5 1lim
2 4x
x xx
2
2
5 13lim 2 4x
x x
x
3 0 0 30 4 4
Divide by 2x
2
2
5 1lim 3 lim lim
2lim lim 4
x x x
x x
x x
x
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More Examples3 2
3 2
2 3 21. lim100 1x
x xx x x
3 2
3 3 3
3 2
3 3 3 3
2 3 2
lim100 1x
x xx x x
x x xx x x x
3
2 3
3 22lim 1 100 11x
x x
x x x
2 21
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0
2
3 2
4 5 212. lim7 5 10 1x
x xx x x
2
3 3 3
3 2
3 3 3 3
4 5 21
lim7 5 10 1x
x xx x x
x x xx x x x
2 3
2 3
4 5 21
lim 5 10 17x
x x x
x x x
0
7
2 2 43. lim12 31x
x xx
2 2 4
lim 12 31x
x xx x xxx x
42lim 3112x
xx
x
212
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24. lim 1x
x x
22
2
1 1 lim1 1x
x x x x
x x
2 2
2
1lim1x
x x
x x
2
1 lim1x x x
1 1 0
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Infinite LimitsFor all n > 0,
1lim nx a x a
1lim if is evennx a
nx a
1lim if is oddnx a
nx a
-8 -6 -4 -2 2
-20
-15
-10
-5
5
10
15
20
-2 2 4 6
-20
-10
10
20
30
40
-8 -6 -4 -2 2
-20
-15
-10
-5
5
10
15
20
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Examples Find the limits
2
20
3 2 11. lim2x
x xx
2
0
2 13= lim
2x
x x
32
3
2 12. lim2 6x
xx
3
2 1= lim2( 3)x
xx
-8 -6 -4 -2 2
-20
20
40
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Limit and Trig Functions
From the graph of trigs functions
( ) sin and ( ) cosf x x g x x
we conclude that they are continuous everywhere
-10 -5 5 10
-1
-0.5
0.5
1
-10 -5 5 10
-1
-0.5
0.5
1
limsin sin and lim cos cosx c x c
x c x c
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Tangent and Secant Tangent and secant are continuous everywhere in their domain, which is the set of all real numbers
3 5 7, , , , 2 2 2 2x
-6 -4 -2 2 4 6
-30
-20
-10
10
20
30
-6 -4 -2 2 4 6
-15
-10
-5
5
10
15
tany x
secy x
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Examples
2
a) lim secx
x
2
b) lim secx
x
32
c) lim tanx
x
3
2
d) lim tanx
x
e) lim cotx
x
32
g) lim cotx
x
32
cos 0lim 0sin 1x
xx
4
f) lim tanx
x
1
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Limit and Exponential Functions
-6 -4 -2 2 4 6
-2
2
4
6
8
10
, 1xy a a
-6 -4 -2 2 4 6
-2
2
4
6
8
10 , 0 1xy a a
The above graph confirm that exponential functions are continuous everywhere.
lim x c
x ca a
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Asymptoteshorizontal asymptotThe line is called a
of the curve ( ) if eihterey L
y f x
lim ( ) or lim ( ) .x x
f x L f x L
vertical asymptote The line is called a of the curve ( ) if eihter
x cy f x
lim ( ) or lim ( ) .x c x c
f x f x
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ExamplesFind the asymptotes of the graphs of the functions
2
2
11. ( )1
xf xx
1 (i) lim ( )
xf x
Therefore the line 1 is a vertical asymptote.
x
1.(iii) lim ( )x
f x
1(ii) lim ( )
xf x
.
Therefore the line 1 is a vertical asymptote.
x
Therefore the line 1 is a horizonatl asymptote.
y
-4 -2 2 4
-10
-7.5
-5
-2.5
2.5
5
7.5
10
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2
12. ( ) 1
xf xx
21 1
1(i) lim ( ) lim 1x x
xf xx
1 1
1 1 1= lim lim .( 1)( 1) 1 2x x
xx x x
Therefore the line 1 is a vertical asympNO t eT ot .
x
1(ii) lim ( ) .
xf x
Therefore the line 1 is a vertical asymptote.
x
(iii) lim ( ) 0.x
f x
Therefore the line 0 is a horizonatl asymptote.
y
-4 -2 2 4
-10
-7.5
-5
-2.5
2.5
5
7.5
10
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Thank You
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