Section 2.7: Precise Definition of Limits

Section 2.7: Precise Definition of Limits
Assume f (x) exists for all x in some open interval containing a, except possibly
at a. We say that the limit of f (x) as x approaches a is L, written
lim f (x) = L
x→a
, if for any number ε > 0, there is a corrseponding number δ > 0 such that
|f (x) − L| < ε whenever 0 < |x − a| < δ
Example: The following graph shows a linear function such that limx→3 f (x) =
5. For each ε > 0 given, find the corresponding δ > 0 satisfying the statement
|f (x) − 5| < ε whenever |x − 3| < δ
a) ε = 1.
Solution: ε = 1 implies that we want to find the interval where, on the graph
satisfies the inequality
|f (x) − 5| < 1 whenever |x − 3| < δ
On the graph, that corressponds to the following picture:
We see that the line intersects the ”ε strip” at the values x = 1 and x = 5.
Since we are looking at when x → 3, we know that it satisfies the inequality
|x − 3| < 2
1
since 5 is 2 bigger than 3, while 1 is 2 smaller than 3. This tells us that
δ≤2
b) ε = 1/2.
Solution: Same as a), but we have a different ε strip:
Giving that
|f (x) − 5| < 1/2 whenever |x − 3| < 1
Giving that δ ≤ 1.
1
Writing Proofs
1. Find δ: Let ε be an arbitrary positive number. Use the inequality |f (x) −
L| < ε to find a condition of the form |x − a| < δ, where δ depends only on the
value of ε.
2. Write a Proof. For any ε > 0, assume 0 < |x − a| < δ and use the relationship between ε and δ in Step 1 to prove that |f (x) − L| < ε.
Example: Write a proof that limx→4 (4x − 15) = 1 usin the precise definition
of a limit.
Solution: In this case, a = 4 and L = 1. Assume ε > 0 is given such that
|(4x − 15) − 1| < ε.
This means that
|4x − 16| < ε
4|x − 4| < ε
ε
|x − 4| <
4
ε
We have found that |(4x − 15) − 1| < ε implies that |x − 4| < . Now that we
4
found the relationship between ε and δ, we can write our proof:
2
Let ε > 0 be given and assume 0 < |x − 4| < δ where δ =
|(4x − 15) − 1| = 4|x − 4| < 4
ε
. Then
4
ε
=ε
4
So, we have shown that for any ε > 0,
|(4x − 15) − 1| < ε whenever |x − 4| < δ
provided 0 < δ ≤ ε/4.
2
Infintie Limits
The infinite limite limx→a f (x) = ∞ means that for any positive number N
there exists a corresponding δ > 0 such that
f (x) > N whenever |x − a| < δ.
Steps for proving that limx→a f (x) = ∞
1. Find δ: Let N be an arbitrary positive number. Use the inequality f (x) > N
to find a condition of the form |x − a| < δ, where δ depends only on the value
of N .
2. Write a Proof. For any N > 0, assume 0 < |x − a| < δ and use the
relationship between N and δ in Step 1 to prove that f (x) > N .
Example: Prove that lim
x→2
1
=∞
(x − 2)2
Solution: Assuming N > 0, we use the inequality
1
>N
(x − 2)2
to find δ only depending on N . This inequality gives us
(x − 2)2 <
1
N
Taking square roots of both sides gives
1
|x − 2| < √
N
1
So, we found our δ = √ that only depends on N . So we can write the proof:
N
1
1
Let N > 0 be given. If |x − 2| < √ , then
> N . This proves that
(x
−
2)2
N
lim
x→2
1
=∞
(x − 2)2
3
3
Homework Problems:
1. Suppose |f (x) − 7| < 0.3 whenever 0 < x < 7. Find all values of δ > 0
such that |f (x) − 7| < 0.3 whenever |x − 2| < δ.
Solution: The original inequality involving x is 0 < x < 7. The question
is asking for when |x − 2| < δ, but we know this holds for 0 < x < 7. So,
we look at the endpoints, and choose the smaller of the 2 distances from
x = 2:
|2 − 0| = 2
|2 − 7| = 5
So, it will hold for all δ ≤ 2.
2. Use the graph below to find the maximum value of δ > 0 such that for all
x |x − 2| < δ implies |f (x) − 4| < ε
√
Solution: We see that it holds for the interval 1 < x < 7, but since
√ we
need the inequality |x − 2| < δ to hold, it will be true for all δ ≤ 7 − 2,
which is the shorter of the two distances between 2 and the endpoints.
3. Give an ε − δ proof of lim (x2 − x − 4) = 2
x→−2
Solution: Given ε > 0, we want to find δ such that
|x2 − x − 4 − 2| < ε whenever |x + 2| < δ
The first part of that statement can factor into
|x + 2||x − 3| < ε
First assume that δ ≤ 1 Then we have that
0 < |x + 2| < 1.
4
Then, we have
|x − 3| = |x + 2 − 5| ≤ |x + 2| + | − 5| = |x + 2| + 5 < 1 + 5 = 6
So, we are left with
6|x + 2| < ε
ε
. Our formal proof becomes:
6
n εo
. Then |x + 2| < δ imples
Let ε > 0 be given. Choose δ = max 1,
6
2
|x − x − 4 − 2| < ε
or that δ =
5