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Unformatted text preview: V is a
neighborhood of . Using condition c we choose a number 0 such that the condition f x
V
will hold whenever x S
a , a
a . Since the interval a , a is a neighborhood of
a, condition a must hold.
7. Given that S is a subset of a metric space X, that f is a function from X into a metric space Y, that
that a is an interior point of S, prove that the following conditions are equivalent:
a. f x as x Y and a. b. For every number 0 there exists a number 0 such that the inequality d f x ,
every point x X
a that satisfies the inequality d x, a .
209 will hold for The only way in which condition b differs from the , form of the assertion that f x
as x a is
that it requires that d f x , for all points x within a distance of a and unequal to a. It does not
merely assert that d f x , when x is a member of the set S unequal to a and within a distance
of a.
It is obvious that condition b implies condition a. To show that condition a implies condition b we
assume that f x
as x a. Suppose that 0. Choose a number 1 0 such that the
condition d f x , will hold whenever x S B a, 1
a . Now, using the fact that a is an
interior point of S, choose a number 2 0 such that B a, 2
S. We define to be the smaller of
the two numbers 1 and 2 and we observe that the inequality d f x , will hold for every point
x that satisfies the inequality d a, x .
8. Suppose that S is a set of real numbers, that a is a limit point of S, that f : S R and that is a real number.
Complete the following sentence: The function f fails to have a limit of at the number a when there exists a
neighborhood V of such that ......
The function f fails to have a limit of at the number a when there exists a number 0 such that
for every number 0 there is at least one number x in the set S
a , a
a for which
.
f x 
9. Prove that
1 x xy
1 x 2 xy y 2
as x, y
1, 2 .
We begin with the observation that if u x
1 x xy
1 x 2 xy y 2 1
4 1
4 1 and v y 2 then 1 u1 u1 v2
1 u1 2 u1 v2 v2 2 1
4 u 2 v 2 v 3uv
1
.
4 8 u 2 4u uv 5v v 2
In the event that x 2  1 and y 2  1 we see that
2
2
1
1
2 4u uv 5v v 2
51
3
8 u
 8 u2 v2 4 1
2
2
2
2
and so
u 2 v 2 v 3uv 
u 2 v 2 v 3uv
1
.
4 8 u 2 4u uv 5v v 2
12
Suppose that 0 and define to be the smaller of the two numbers 1 and . Then whenever
2
1, 2 , defining, u x 1 and v y 2, we have
x, y
1, 2 and x, y
u 2 v 2 v 3uv 
1 x xy
1
6 .
2 xy y 2
4
12
12
1x
10. In each of the following cases, determine whether or not the function f has a limit at 0, 0 . Use
Scientific Notebook to view the graph of the function f in each case.
a. For each point x, y 0, 0 we define
f x, y 210 x2y
.
y2 x2 2
0
2 y
2 2 x 4 4 4 Since
x2
1
x2 y2
whenever x, y
0, 0 we have f x, y  y  whenever x, y
0, 0 . Now we can show that
f x, y
0 as x, y
0, 0 . Suppose that 0. We define an...
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 Fall '08
 STAFF
 Math, Calculus

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