Section 1.3: Limits
January 31st , 2017
MATH 150A: Lecture #6 Notes
Topics Covered
Page
1 Intuitive Definition of a Limit
1
Example 1
2
2 One-Sided Limits
3
Example 1
4
Example 2
4
Example 3
5
3 Precise definition of a limit
5
Example 1
6
Example 2
6
1 Intuitive Definition of a Limit
Consider the function f (x) = x2 − x + 2. The graph of f is given by:
WE have this
x
f (x)
x
1
1.5
1.8
1.9
1.95
1.99
1.995
1.999
f (x)
3
2.5
2.2
2.1
2.05
2.01
2.005
2.0001
Table 1: From the right
Table 2: From the left
When x is close to 2, on either side, f (x) is close to
.
We can see from these tables that when x is close to 2 (on either side of 2), the value of f (x) is close to 4. In
fact, we can make the values of f (x) as close as we like to 4 by taking x sufficiently close to 2. This means
that
lim f (x) = 4.
x→2
Question: What if x = 2 is not in the domain as in f (x) =
1 Intuitive Definition of a Limit continued on next page. . .
(x−2)(x2 −x+2)
?
x−2
Page 1 of 6
Section 1.3: Limits
MATH 150A: Lecture #6 Notes
January 31st , 2017
Answer:
Example 1
#1
Remark 0.1.
The limit of the following functions are the same:
Figure 1: lim f (x) in 3 different scenarios.
x→L
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Section 1.3: Limits
January 31st , 2017
MATH 150A: Lecture #6 Notes
2 One-Sided Limits
Let
(
H(t) =
0
1
if t < 0
if t > 0
The function H(t) is called Heaviside function.
• From the left, lim− H(t) = 0 (by considering values of H(t) when t < 0).
t→0
• From the right, lim H(t) = 1 (by considering values of H(t) when t > 0).
t→0+
Hence,
lim H(t)
t→0−
does not exist
Figure 2: Graph of H(t).
If lim− f (x) 6= lim+ f (x), then lim f (x) does not exist.
x→a
#1
x→a
x→a
In other word,
lim f (x) = L if and only if
x→a
lim f (x) = L and
x→a−
lim f (x) = L
x→a+
Page 3 of 6
Section 1.3: Limits
MATH 150A: Lecture #6 Notes
January 31st , 2017
Example 1
π
Find lim sin
x→0
x
+ The function x → sin πx is defined whenever x 6= 0.
+ Evaluate f (x) for some small values of x. Particularly,
•f (1) =
•f ( 12 ) =
•f ( 31 ) =
•f ( 14 ) =
We might guess that lim sin
x→0
π
=
x
#2
Example 2
1
if it exists.
x→0 x2
+ If x becomes close to 0, the x2 becomes also close to 0 and
Find lim
x
1
x2
becomes very large. Indeed,
f (x)
±1
± 0.5
± 0.1
± 0.05
±0.001
±0.0001
Table 3: Values of f (x) when x → 0, from both
sides.
#3
Figure 3: Graph of f (x) =
1
x2 .
Page 4 of 6
Section 1.3: Limits
January 31st , 2017
MATH 150A: Lecture #6 Notes
Remark 0.2. The 3 following statement are equivalent:
i) f (x) does not have equal one-sided limits,
ii) Does not approach anything,
iii) Infinit limit.
Example 3
The following graph represents the function g. Find the following limits:
• lim f (x) =
x→1
• lim− f (x) =
x→3
#4
• lim+ f (x) =
x→3
• lim f (x) =
x→3
Figure 4: Graph of f (x).
•f (3) =
3 Precise definition of a limit
Precise Definition of a Limit: Let f be a function defined on some open interval that contains the number
a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write
lim f (x) = L
x→a
if for every number > 0 there is a corresponding number δ > 0 such that if 0 < |x−a| < δ then |f (x)−L| < .
− δ Proofs for Limits: Overall Approach:
1. Let > 0 be arbitrarily chosen.
#1
2. Find a δ for that that will make the inequality |f (x) − L| < true.
3. Rewrite it in proof form.
Page 5 of 6
Section 1.3: Limits
January 31st , 2017
MATH 150A: Lecture #6 Notes
Example 1
Prove that
lim (4x − 5) = 7.
x→3
Steps: (Scratch work prior to proof )
1. First determine f (x) = 4x − 5, a = 3, L = 7. Then we need to find a value for δ.
2. The short-term objective is to obtain the form |x − 3| <(something).
3. Work backwards, starting with what we want to show:
|(4x − 5) − 7| < 4. Simplify the inequality, trying to obtain |x − 3|:
|4x − 12| < ⇒ |4||x − 3| < ⇒ |x − 3| <
4
5. Use this (something) as the δ, i.e. δ = 4 .
Proof:
#2
• Let > 0.
(Always the first line. This indicates that we can get arbitrarily close to L.)
• Let δ = 4
(In the definition, we must show there is a δ for each . So we must give a δ that will work. This is
taken from the preliminary work.)
• Then for every x such that 0 < |x − 3| < δ, we have 0 < |x − 3| < 4 .
(The first part is wording from the definition. The second part is just a substitution to be used later.)
• Thus, |(4x − 5) − 7| = |4x − 12| = |4||x − 3| = 4|x − 3| < 4 4 = .
(We rewrite the steps from the scratch work, and then substitute in the inequality for |x − 3| < 4 . )
• Then by the definition of the limit,
lim (4x − 5) = 7.
x→3
(We have shown that the requirements for the definition have been satisfied.)
Example 2
Prove these statements using the ε, δ definition of a limit:
2 + 4x
=2
x→1
3
lim
#3
4x
3−
= −5.
x→10
5
and lim
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