Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN: 9781133382119
Author: Swokowski
Publisher: Cengage
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Please answer 4.2.1 (a,b)
![Corollary 4.2.4 (Algebraic Limit Theorem for Functional Limits). Let
f and g be functions defined on a domain ACR, and assume lime f(x) = L
and lime 9(x)=M for some limit point c of A. Then,
(i) lim kf(x)
LC
=
kL for all kЄ R,
(ii) lim [f(x)+9(a)] = L+M,
C
(iii) lim [f(x)g(x)] = LM, and
91F
(iv) lim f(x)/9(x)=L/M, provided M +0.
Theorem 4.2.3 (Sequential Criterion for Functional Limits). Given a
function f: AR and a limit point c of A, the following two statements are
equivalent:
(i) lim f(x) = L.
01-2
(ii) For all sequences (an) C A satisfying xn c and (xn)c, it follows that
f(xn) → L.
23 The Algebraic and Order Limit Theorems
(i) To prove this statement, we need to argue that the quantity
(an+bn)-(a+b)|
51
can be made less than an arbitrary e using the assumptions that (an-al and
- can be made as small as we like for large n. The first step is to use the
triangle inequality (Example 1.2.5) to say
(an+bn)-(a+b)| = |(ana) + (bnb)|≤lan-al + lb-bl.
Again, we let >0 be arbitrary. The technique this time is to divide the e
between the two expressions on the right-hand side in the preceding inequality.
Using the hypothesis that (an) a, we know there exists an N₁ such that
lan-al<
whenever n≥ N₁.
Likewise, the assumption that (bn) b means that we can choose an N₂ so
|bb|< whenever n≥ N₂.
that
-
The question now arises as to which of N₁ or N₂ we should take to be our
choice of N. By choosing N = max(N1, N2), we ensure that if n ≥ N, then
> N₁ and n ≥ №2. This allows us to conclude that
(an+bn)-(a+b)| ≤an-a|+|bn-b
€
for all n≥N, as desired.
< + = €
2](https://content.bartleby.com/qna-images/question/c6389447-1237-4af0-b5c6-eb1260425b55/57037c49-1a57-4687-a59a-c98a0bfa9828/bdbgjp69_thumbnail.png)
Transcribed Image Text:Corollary 4.2.4 (Algebraic Limit Theorem for Functional Limits). Let
f and g be functions defined on a domain ACR, and assume lime f(x) = L
and lime 9(x)=M for some limit point c of A. Then,
(i) lim kf(x)
LC
=
kL for all kЄ R,
(ii) lim [f(x)+9(a)] = L+M,
C
(iii) lim [f(x)g(x)] = LM, and
91F
(iv) lim f(x)/9(x)=L/M, provided M +0.
Theorem 4.2.3 (Sequential Criterion for Functional Limits). Given a
function f: AR and a limit point c of A, the following two statements are
equivalent:
(i) lim f(x) = L.
01-2
(ii) For all sequences (an) C A satisfying xn c and (xn)c, it follows that
f(xn) → L.
23 The Algebraic and Order Limit Theorems
(i) To prove this statement, we need to argue that the quantity
(an+bn)-(a+b)|
51
can be made less than an arbitrary e using the assumptions that (an-al and
- can be made as small as we like for large n. The first step is to use the
triangle inequality (Example 1.2.5) to say
(an+bn)-(a+b)| = |(ana) + (bnb)|≤lan-al + lb-bl.
Again, we let >0 be arbitrary. The technique this time is to divide the e
between the two expressions on the right-hand side in the preceding inequality.
Using the hypothesis that (an) a, we know there exists an N₁ such that
lan-al<
whenever n≥ N₁.
Likewise, the assumption that (bn) b means that we can choose an N₂ so
|bb|< whenever n≥ N₂.
that
-
The question now arises as to which of N₁ or N₂ we should take to be our
choice of N. By choosing N = max(N1, N2), we ensure that if n ≥ N, then
> N₁ and n ≥ №2. This allows us to conclude that
(an+bn)-(a+b)| ≤an-a|+|bn-b
€
for all n≥N, as desired.
< + = €
2

Transcribed Image Text:Exercise 4.2.1.
(a) Supply the details for how Corollary 4.2.4 part (ii) follows
from the Sequential Criterion for Functional Limits in Theorem 4.2.3 and
the Algebraic Limit Theorem for sequences proved in Chapter 2.
(b) Now, write another proof of Corollary 4.2.4 part (ii) directly from Defini-
tion 4.2.1 without using the sequential criterion in Theorem 4.2.3.
(c) Repeat (a) and (b) for Corollary 4.2.4 part (iii).
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