Math 414
Professor Lieberman
March 24, 2003
SOLUTIONS TO SECOND INCLASS EXAM
1. (a) Let M > 0 be given and choose δ = 1 /M . If 0 < x
−
1 < δ , then 1 < x < 1 + δ , so x x
−
1
≥
1
>
1
δ
= M.
x
−
1
(b) Let ε > 0 be given and choose δ = min { 1 , ε } . If 0 <  x − 2  < δ , then 1 < x < 3, so x x + 2
<
3
3
= 1 , and therefore x
2
+ 4
−
2 x + 2
= x x
+ 2
 x
−
2

<
 x
−
2

< ε.
(c) Let ε > 0 be given and choose M = 6 /ε . If x <
−
M (and x is irrational), then
 f ( x )
−
2

=
6
3
− x
<
6
3 + M
<
6
M
= ε.
2. (a) (Remember that f is must be defined on all of the interval [0 , 1].) One function that works f ( x ) =
(
2 x
2 x
−
1 if 0 if 1 /
≤
2 x
≤
1
< x
≤
/ 2
1 .
,
(b) The easy example to write down is Dirichlet’s function on [0 , 1]: f ( x ) =
(
1 q
0 if x = p q in lowest terms otherwise .
,
3. (a) This one is false. Take f ( x ) = x and g ( x ) = sin(1 /x ). Then lim x
→
0
( f g )( x ) = lim x
→
0 f ( x ) = 0 , but lim x
→
0 g ( x ) does not exist; however, if the two limits exist, then one of them must be zero.
(b) Set f ( x ) = 2 x
−
3 x . Then f (0) = 1 and f (1) =
−
1, so f ( x ) must be zero at some with 0 < c < 1 by the intermediate value theorem. But when x = c , it follows that 2 x x = c
= 3 x .
1