# Math 414 Professor Lieberman March 24, 2003 SOLUTIONS TO SECOND IN-CLASS EXAM Math 414

Professor Lieberman

March 24, 2003

SOLUTIONS TO SECOND IN-CLASS 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