Download Section 2.4 1 Definition of a Limit 2 The Absolute Value Function

Section 2.4
So far we have been using a lot of geometry to give us proper intuition for
taking limits, however to move much further we need to establish a solid
theory where as we can talk about limits exactly and not just as some thing
that appears on a graph.
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Definition of a Limit
Here we are going to accurately define what we mean by ”near” and how
nearness realtes to the limit of a function as we discussed geometrically.
Definition:[Limit] Let f be a function defined on the open interval
(c − p, c + p) except perhaps at c itself. We say limx→c f (x) = L, if for all
ε > 0, then there exists a δ > 0 such that whenever |x − c| < δ then
|f (x) − L| < ε.
Let us explore what is going on here, we are saying that for any ε > 0,
this means ε could be .5, .001, 1, 3000, .0000000001919191878186837634876,
whatever, that is no matter how small we make ε then there is also some small
δ, such that when the difference of c and x is within δ then the didderence
from f (x) and L is with in ε. This delivers for us the notion of ”near” that I
have up to this point not explained. We will be using this definition to prove
that the limits we calculate are indeed the limits of the function. In fact, we
can use this to prove some of our limit properties. But first I want us to go
over the absolute value funtion.
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The Absolute Value Function
The absolute value function | ? | : R → R is defined by
(
a
if a ≥ 0
|a| =
−a if a < 0
Theorem:[The Triangle Inequality] Let a and b be real numbers, then
|a + b| ≤ |a| + |b|.
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Proof: As a preliminary step observe that for any real number r
we have the inequality −|r| ≤ r ≤ |r|. Hence, −|a| ≤ a ≤ |a| and
−|b| ≤ b ≤ |b|. Adding these two inequalties of a and b give us
that −(|a| + |b|) ≤ a + b ≤ |a| + |b|. That is −(|a| + |b|) ≤ a + b,
giving us the inequality as −(a+b) ≤ |a|+|b|, and a+b ≤ |a|+|b|.
Since |a + b| is either a + b or −(a + b) we conclude that |a + b| ≤
|a| + |b|.
Q.E.D.
Theorem: Let a and b be real numbers, then |ab| = |a||b|.
Proof: This proof considers 4 cases.
Case 1: Suppose that a ≥ 0 and b ≥ 0, then |ab| = ab = |a||b|.
Case 2: Suppose that a ≤ 0 and b ≤ 0, then −a ≥ 0 and −b ≥ 0,
hence |a||b| = (−a)(−b) = ab = |ab|
Case 3: Suppose that a ≥ 0 and b ≤ 0, then −b ≥ 0, so |a||b| =
a(−b) = −ab = |ab|.
Case 4: Suppose that a ≤ 0 and b ≥ 0, the by a similar argument
as in case 3 we get that |a||b| = |ab|
Q.E.D.
The previous theorems are nice tricks that will come up to prove limits
exist.
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Examples of Proving the Existance of a Limit
with the ε − δ Definition
Example 1) Evaluate the following limit limx→3 x2 and prove that your evaluation satisfies the ε − δ definition.
Proof: First we wish to evaluate the limit, which is this case is
easy enough, it should be 9. We now prove that the limit is infact
9 as we approach 3. The begining of these proofs are all the same
”Let ε > 0” be given. Once we declare our ε then we begin to
find a δ that bounds our x values. We need to show that for
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our choice of δ that our ε bounds f and L the way the definition
states it should.
Discussion: Our goal is for |x2 − 9| < ε, but |x − 3||x + 3| =
|x2 − 9|. We want |x − 3| < δ, so we need to deal with this
|x + 3| term. Suppose that |x − 3| < 1, then −1 < x − 3 < 1,
so 2 < x < 4. That is x can be no larger then 4. Now back to
|x + 3|. By the triangle inequality, |x + 3| ≤ |x| + |3| but x is less
than 4, hence |x| + |3| < 4 + |3| = 7. What we want then is for
7|x − 3| < ε, by this condition we can then let δ = ε/7 which is
a positive real number and we’re done.
This is what the proof should look like Let ε > 0 be given.
Then for all x where 0 < |x − 3| < δ = min (ε/7, 1) we have that
|x2 − 9| = |x − 3||x + 3| < δ7 = 7ε 7 = ε.
Q.E.D.
Example 2) Prove that limx→4 x1/2 = 2
Proof:
Discussion: We want for |x1/2 − 2| < ε, regardless of what
epsilon we choose. So notice that |x − 4| = |x1/2 − 2||x1/2 + 2|.
Recall, that x1/2 is defined for values greater than or equal to 0,
this means that we can force our δ to be less than or equal to 4
and it also suggests that |x1/2 + 2| ≥ 2, hence |x1/2 − 2| < |x − 4|.
Let δ = min (ε, 4). Then we can write the proof.
Formal Proof: Let ε > 0 be given and take 0 < δ = min (ε, 4).
Then whenever |x − 4| < δ we have that |x1/2 − 2| < |x1/2 −
2||x1/2 + 2| = |x − 4| < ε.
Q.E.D.
As a good measure you should provide the discussion and the formal proof
when you encounter these types of problem. Given this information we can
now prove all of the Algebraic Properties from section 2.3.
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