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Unformatted text preview: Now we want to show that the function f has no limit at the number 2. To obtain a contradiction, suppose
that is a limit of the function f at 2. The key to the desired contradiction is the fact that when x 2 we
have f x x 2 which is close to 4 when x is close to 2 and when x 2 we have f x x which is close to 2
when x is close to 2. We shall use this observation to argue that both of the numbers 2 and 4 must lie close
to the limit value , in spite of the fact that the distance from 2 to 4 is 2.
Using the fact that 1 0 and that is a limit of f at 2, we choose a number 0 such that the inequality
f x  1
holds whenever x 2 and x 2  . Choose a number x 1 2 and a number x 2 2 such that
x 1 2  and x 2 2  . Then
f x 2  f x 1
f x2 
f x 1
  f x 2  1 1 2.
f x 1
2
On the other hand, f x 1 x 1 2 and f x 2 x 2 4 which gives us
f x 2  2.
f x 1
We have therefore reached the desired contradiction.
3. Given that
fx
prove that f x 1 as x x if x is rational x 2 if x is irrational 1 and prove that this function f has no limit at the number 2. Solution: This solution is very similar to the solution of Exercise 2 but we have to exercise a
little more care with the inequalities at the end.
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 2 and
x 1  /3 then, regardless of whether f x 1  x 1  or f x 1  x 2 1  we have
.
3
With this fact in mind we define /3 and we observe that f x 1 
domain of f and x 1 and x 1  .
f x Note that we also have the inequality f x 1 3x 1 1 3 whenever x lies in the when x 1 but we do not need this fact. Now we want to show that the function f has no limit at the number 2. To obtain a contradiction,
suppose that is a limit of the function f at 2. The key to the desired contradiction is the fact that
when x is irrational we have f x x 2 which is close to 4 when x is close to 2, and when x is rational
we have f x x which is close to 2 when x is close to 2. We shall use this observation to argue 207 that both of the numbers 2 and 4 must lie close to the limit value , in spite of the fact that the
distance from 2 to 4 is 2.
Using the fact that 1 0 and that is a limit of f at 2, we choose a number 0 such that the
inequalities
f x  1
2
1
2
4
x
2
1
x 2 
2
all hold whenever x 2 and x 2  . Choose a rational number x 1 and an irrational number x 2
such that x 1 2  and x 2 2  . Then
f x 2  f x 1
f x2 
f x 1
  f x 2  1 1 1.
f x 1
2
2
On the other hand, f x 1 x 1 and f x 2 x 2 2 and so
f x 2  f x 2
2 1 1 1 3
4 2 4 f x 2  f x 1
2
2
2
2
f x 2  2.
f x 1
and we have therefore reached the desired contradiction.
4. Given that for every number x 2
9
fx x
x 3 
3, prove that f has no limit at the...
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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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