Unformatted text preview: x 1
5x 2 4 3
5 7. Given that
fx 1 if x is rational
0 if x is irrational explain why f does not have a limit from the right at 2. Solution: To obtain a contradiction, suppose that is a limit of the function f from the right at 2. Using the fact that 1/2 0 we choose a number 0 such that the inequality
f x  1
2
will hold for every number x
2, 2 . Choose a rational number x and an irrational number t in the
interval x, x . Then
1 f x f t  f x   f t  1 1 1
2
2
and we have reached the desired contradiction. 8. Suppose that a is an interior point of a set S of real numbers and that f : S R. Suppose that f x
0 as
x a and that f x
1 as x a . Prove that the function f does not have a limit at the number a.
Since x must be a limit point of each of the sets Ý, a
S and S
a, Ý , the desired result follows
at once from Theorem 8.3.2. Some Exercises on Continuity
1. Given that
f x x2 1
x 3 191 for every number x, prove that the function f is continuous at the number 2.
1
All we have to show is that f x
as x 2. This fact can be deduced directly in the same way
7
that we did the earlier exercises on limits or, if you prefer, it can be deduced at once from
Theorem 8.5.4.
2. Given that
x sin fx 1
x if x 0 if x 0 0 prove that the function f is continuous at the number 0. Hint: Use the fact that f x  x  for every number x
and use the sandwich theorem. The graph of this function is illustrated in the following figure. From the inequality
0 f x f 0  x 
that holds for every number x and from the sandwich theorem we see at once that f x
x 0. f 0 as 3. Given that f is the ruler function defined earlier, explain why f is continuous at every irrational number in the
interval 0, 1 and discontinuous at every rational number in 0, 1 . Solution: Since the equation f x 0 holds for x
0, 1 if and only if the number x is irrational, all
we have to do is show that for every number x
0, 1 we have
lim f t 0.
tx Suppose that x
0, 1 and suppose that 0. Choose an integer N 1/ . Since there are only finitely
many rational numbers that have a reduced form m/n for which n N we know that there are at most
finitely many numbers x in the interval 0, 1 for which f x
. We shall call this finite set S. Since the set
S
x , being finite, is closed, the set
R
S
x
must be open and is therefore a neighborhood of x. We also know that whenever
t 0, 1 0 as t f t
x. R S x we have
and so we have shown that f t 0 f t 4. Suppose that f and g are functions from a given set S of real numbers into R and that the inequality
f t f x  g t g x 
holds for all numbers t and x in S. Prove that f must be continuous at every number at which the function g is
continuous.
Suppose that the function g is continuous at a given number x. Suppose that 0. Choose 0
such that the inequality g t g x  will hold whenever t S
x , x . Then for every such
number t we have 192 f t fx  gx  . g t 5. Given that f is a continuous function...
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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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