1.2 finding limits graphically and numerically
DESCRIPTION
Calculus. 1.2 Finding Limits Graphically and Numerically. Numerical Investigation. Create a table for x values approaching 0. Graphic Investigation. Find the limit of f(x) as x approaches 2 where f is defined as. Non-existance. Example 3:. - PowerPoint PPT PresentationTRANSCRIPT
1.2 Finding Limits Graphically 1.2 Finding Limits Graphically and Numericallyand Numerically
xx -0.01-0.01 -0.001-0.001 -0.0001-0.0001 00 0.00010.0001 0.0010.001 0.010.01
f(x)f(x) 1.99501.9950 1.99951.9995 1.99991.9999 ?? 2.00012.0001 2.00052.0005 2.00502.0050
Create a table for x values approaching 0
0
limxf x
0lim
1 1x
x
x
2
Find the limit of f(x) as x approaches 2 where Find the limit of f(x) as x approaches 2 where ff is defined as is defined as
2
1
2
limx
f x
1
f x 1, 20, 2{ xx
Example 3:Example 3:
Solution: Consider the graph of the function f(x) = |x|/x.
|x|/x =1, x>0
1
1
|x|/x =-1, x<0 0limx
x
xDNE
Non-existance
0limx
x
x
Example 4:Example 4:
Discuss the existence of the limit:Discuss the existence of the limit:
Solution: Using a graphical representation, you can see that x does not approach any number. Therefore, the limit
20
1limx x
20
1limx x
20
1limx x
DNE
.DNE
Non-existance
20
1limx x
Example 5:Example 5:
xx xx→0→0
sin 1/xsin 1/x 11 -1-1 11 -1-1 11 -1-1 DNEDNE
Make a table approaching 0
The graph oscillates, so no limit exists.
2 / 2 / 3 2 / 5 2 / 7 2 / 9 2 / 11
Non-existance
0
1limsinx x
DNE
0
1limsinx x
Common types of behavior associated with Common types of behavior associated with nonexistence of a limitnonexistence of a limit
1) f(x) approaches a different number from the right side of c than it approaches from the left side.
2) f(x) increases or decreases without bound as x approaches c.
3) f(x) oscillates between two fixed values as x approaches c.
Homework
Definition of LimitDefinition of Limit
Let f be a function defined on an open interval Let f be a function defined on an open interval containing c (except possibly at c) and let L containing c (except possibly at c) and let L be a real number. The statement:be a real number. The statement:
lim f(x) = Llim f(x) = LXX→c→c
Means that for each > 0 there exists a >0 such that if 0 < |x-c| < then |f(x)-L|<
Means that for each > 0 there exists a >0 such that if 0 < |x-c| < then |f(x)-L|<
1
1
1
lim 1xf x
f x x
|x-c|
|f(x)-L|
Ex. 7
Ex. 8
Example 7:Example 7:
Prove:Prove: Lim (Lim (3x-23x-2) = ) = 44XX→2→2
Solution: You must show that for each > 0, there exists a >0 such that |(3x-2)-4)| < . Because your choice of depends on , you need to establish a connection between
|(3x-2)-4| and |x-2|.
|(3x-2)-4| = |3x-6| = 3|x-2|
Thus, for a given > 0 you can choose 3. This choice works because
0< |x-2| < = /3
|(3x-2)-4| = 3|x-2| < 3(/3) =
"The distance between and 2
is less than delta and
delta = epsilon over 3".
x
Diagram
Example 8:Example 8:
Use the Use the -- definition of a limit to prove that: definition of a limit to prove that:
Lim xLim x22 = 4 = 4 XX→2→2
Solution: You must show that for each > 0, there exists a > 0 such that
HW 1.2 p. 53/3-40 (Just find the limits), 42-46
when 0 < |x-2| < |x2 - 4| <
2 4 2 2
in 1,3 a neighborhood of 2
2 5
x x x
x
x
So, 0 < |x-2| < = /5
2 4 2 2 5 2 5 / 5x x x x
Diagram
Created by Brandon Marsh and Sean Beireis.Created by Brandon Marsh and Sean Beireis.
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