This preview shows page 1. Sign up to view the full content.
Unformatted text preview: e that
x fx 3 if x 3 x3 if x 3 To obtain a contradiction, suppose that the function f has a limit at 3. Choose 0 such that
1 and such that whenever x 3  and x 3 we have
f x  1.
Choose numbers x 1 and x 2 such that
3 x 1 3 x 2 3 .
We observe that
f x 2  f x 1
 f x 2
 1 1 2.
f x 1
On the other hand
f x1 x1 3 2 3 5
and
f x2 x2 3 6
and so
f x2  6
5 2.
f x 1
This is the desired contradiction.
5. Given that S is a set of real numbers, that f : S R, that is a real number and that a is a limit point of S,
prove that the following conditions are equivalent:
a. f x as x a. b. For every number 0 there exists a number 0 such that the inequality f x
every number x in the set S
a that satisfies the inequality x a  .  3 holds for c. For every number 0 there exists a number 0 such that the inequality f x
every number x in the set S
a that satisfies the inequality x a  5.  3 holds for Solution: We shall provide the proof that condition c implies condition a. Suppose that condition c
holds. To prove that condition a holds, suppose that 0. Using the fact that the number /3 is positive
we now apply condition c to choose a number 0 such that the inequality
f x holds whenever x
x a  . S a and x  3 3
a  5. We see that f x  whenever x S a and 6. Given that S is a set of real numbers, that f : S R, that is a real number and that a is a limit point of S,
prove that the following conditions are equivalent: 187 a. f x as x a. b. For every number 0 there exists a neighborhood U of the number a such that the inequality
a.
f x  holds for every number x in the set U S
c. For every neighborhood V of the number there exists a number 0 such that the condition f x
holds for every number x in the set S
a that satisfies the inequality x a  . V To show that condition a implies condition b we assume that f x
as x a. Suppose that 0.
From condition a and the fact that the interval , is a neighborhood of we deduce that
there exists a neighborhood U of a such that the condition f x
, holds whenever
xUS
a.
To show that condition b implies condition a we assume that condition b holds. Suppose that V is a
neighborhood of . Choose a number 0 such that ,
V. Now, using condition b,
choose a neighborhood U of a such that the condition f x
, holds whenever
xUS
a . Then, whenever x U S
a we have f x
V.
To show that condition a implies condition c we assume that condition a holds. Suppose that V is a
neighborhood of and, using condition a, choose a neighborhood U of a such that the condition
fx
V will hold whenever x U S
a . Choose 0 such that a , a
U. We observe
that whenever x S
a and x a  , we must have x U S
a and so f x
V.
To show that condition c implies condition a we assume that condition c holds. Suppose that V is a
neighborhood of . Using condition c we choose a number 0 suc...
View
Full
Document
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details