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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