Transcript 1.2 Finding Limits Graphically and Numerically
1.2 Finding Limits Graphically and Numerically
x
lim 0
x x
Create a table for x values approaching 0 x f(x) -0.01
1.9950
-0.001
1.9995
-0.0001
1.9999
0 0.0001
? 2.0001
0.001
2.0005
0.01
2.0050
2
x
lim 0
Find the limit of f(x) as x approaches 2 where
f
is defined as { 1,
x
0,
x
2 2 1 2
x
lim 2 1
Example 3:
x
lim 0
x
x
Solution: Consider the graph of the function f(x) = |x|/x.
1 1 |x|/x =1, x>0 |x|/x =-1, x<0
x
lim 0
x x
DNE
Example 4: Discuss the existence of the limit:
x
lim 0
x
1 2 Solution: Using a graphical representation, you can see that x does not approach any number. Therefore, the limit
DNE
.
1
x
lim 0
x
2
x
lim 0 1
x
2
x
lim 0 1
x
2 DNE
Example 5:
x
lim sin
0
1
Make a table approaching 0 x 2 / sin 1/x 1 -1 1 -1 1 -1 x →0 DNE The graph oscillates, so no limit exists.
x
lim sin
0
1
DNE
Common types of behavior associated with nonexistence 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 Limit Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. The statement: lim f(x) = L
X →c
Means that for each
e
> 0 there exists a
d
>0 such that if 0 < |x-c| <
d,
then |f(x)-L|<
e
Means that for each e > 0 there exists a d >0 such that if 0 < |x-c| < d, then |f(x)-L|< e |f(x)-L|
x
e 1 |x-c| 1 d lim
x
1 1
Example 7: Prove: Lim ( 3x-2 ) = 4
X →2
Thus, for a given e > 0 you can choose d e/ 3. This choice works because
0< |x-2| <
d
=
e
/3
Solution: You must show that for each e > 0, there exists a d >0 such that |(3x-2)-4)| < e . Because your choice of d depends on e , you need to establish a connection between is less than delta and delta = epsilon over 3".
|(3x-2)-4| = 3|x-2| < 3( e /3) = e |(3x-2)-4| and |x-2|.
|(3x-2)-4| = |3x-6| = 3|x-2|
Example 8: Use the e d definition of a limit to prove that:
x
2 Lim x 2
X →2
= 4
x
2 Solution: You must show that for each e 0, there exists a d > 0 such that >
when 0 < |x-2| <
d,
|x 2 - 4| <
e .
x
2 a neighborhood of 2
x
x
2 5
So, 0 < |x-2| <
d
=
e
/5
x
2
x
5
x
5 e / 5 e HW 1.2 p. 53/3-40 (Just find the limits), 42-46
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