1.2 Finding Limits Graphically
and Numerically
x
lim

x 0
x  1 1
Create a table for x values approaching 0
x
-0.01
-0.001
-0.0001
f(x)
1.9950 1.9995 1.9999
0 0.0001 0.001
0.01
? 2.0001 2.0005 2.0050
lim f  x   2
x 0
Find the limit of f(x) as x approaches 2 where f
is defined as
f  x  {
1, x  2
0, x  2
1
2
lim f  x   1
x 2
Non-existance
Example 3:
x
lim 
x 0 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  DNE
x 0 x
Non-existance
Example 4:
Discuss the existence of the limit:
1
lim 2 
x0 x
Solution: Using a graphical
representation, you can see
that x does not approach any
number. Therefore, the limit
DNE.
1
1
1
lim 2   lim 2    lim 2    DNE
x 0 x
x0 x
x0 x
Non-existance
Example 5:
1
limsin   
x 0
 x
Make a table approaching 0
2 /  2/ 3  2/ 5  2/  7  2/ 9  2 / 11  x→0
x
sin 1/x 1 -1
1
-1
1
-1
DNE
The graph oscillates,
so no limit exists.
1
limsin    DNE
x 0
 x
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|
1
f  x  x
e
|x-c|
d
1
lim f  x   1
x 1
Ex. 7
Ex. 8
Example 7:
Prove: Lim (3x-2) = 4
X→2
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
Thus, for a given e > 0 you can
choose d  e/3. This choice works
because
0< |x-2| < d = e/3
"The distance between x and 2
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|
Diagram
Example 8:
Use the e-d definition of a limit to prove that:
Lim x2 = 4
X→2
Solution: You must show that for each e >
0, there exists a d > 0 such that
x 4  x2 x2
2
when 0 < |x-2| < d, |x2 - 4| < e .
 x in 1,3  a neighborhood of 2
x2 5
So, 0 < |x-2| < d = e/5
 x2  4  x  2 x  2  5 x  2  5 e / 5  e
HW 1.2 p. 53/3-40 (Just find the limits),
42-46
Diagram
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