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Unformatted text preview: se was upgraded to a theorem. I
have left in the exercise. Sometimes I find it interesting to see which of my students recognize that
an item is the same as one they have already seen. Alt. 8 Limits and Continuity in Metric Spaces
Some Exercises on Limits of Functions
1. Write careful proofs of each of the following assertions:
a. x 3 3x 2 as x We observe 1.
that if x is any given number we have
 x 3 3x 2  x 3 3x 2   x 1 x2 x 2 
The idea of the proof is to make the observation that the factor x 1 will be small when x is
close to 1 and that the factor x 2 x 2 is not too large. In fact, if x 1 1 then, since
2 x 0 we have
x 2 x 2  2 2 2 2 8
which gives us 205  x 3 3x 2   x 1 x 2 x 2  8x 1.
Now suppose that 0. We define to be the smaller of the two numbers 1 and /8. Then,
whenever x 1 we have
 x3 3x 2  x 1 x2 x 2 8x 1 8 8 . We conclude that the inequality
holds whenever x
x 1).
b. c.  x 3 3x 2 
1 and x 1 (and, as a matter of fact, the inequality is also true when 1 as x 3.
3
We begin with the observation that if x is any nonzero number we have
1 3 x  .
1
x
3
3x 
To keep the denominator of this fraction from becoming too small we need to keep x away from
0. In fact, if x 3  1 then, since 2 x 4 we have
1 3 x  3 x  .
1
x
3
6
3x 
Now suppose that 0. We define to be the smaller of the two numbers 1 and and we
observe that whenever x 3  we have
1 3 x  3 x 
1
.
x
3
6
6
3x 
We conclude that the inequality
1
1
x
3
holds whenever x 3 and x 3  (and, as a matter of fact, the inequality is also true when
x 3).
1
x x3 8
12
as x 2.
5
x 6
We begin by observing that if x is any number for which x 2 x 6
2
x3 8
12 x 2 x 2x 4
2x
5
x 2 x3
6
x
x2 0 then we have
12
5 12
x 2 2x 4
5
x3
x 2 5x 8
5 x3
In order to keep the denominator of this fraction from becoming too small we need to keep x
away from 3 and, in order to keep 5x 8 from becoming too large we need to keep x from
being too large. In fact, if x 2 and x 2  1 we have
x 2 5 3 8
x3 8
12 x 2 5x 8
x 2 .
5
5 x3
5 33
x2 x 6
Now suppose that 0 and define to be the smaller of the two numbers 1 and . We see that
whenever x 2 and x 2  we have
x3 8
12 x 2  .
2x
5
6
x
2. Given that
fx x if 0 x 2 x 2 if x 2 206 prove that f x 1 as x 1 and that this function f has no limit at the number 2. Solution: Before we prove that f x 1 as x 1 we make the observation that whenever x 1 1 we have
x 2 1  x 1 x 1 3x 1 .
To prove that f x
1 as x 1, suppose that 0. We observe that whenever x
then, regardless of whether f x 1  x 1  or f x 1  x 2 1  we have
f x 1 3x 1 3 3
With this fact in mind we define /3 and we observe that f x
of f and x 1 and x 1  .
1 Note that we also have the inequality f x 2 and x 1  /3 .
1 whenever x lies in the domain when x 1 but we do not need this fact....
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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