# 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 x − 1

>

1

= M.

δ

(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 must be defined on all of the interval [0 , 1].) One function that works is

(

2 x f ( x ) = if 0 ≤ x ≤ 1 / 2 ,

2 x − 1 if 1 / 2 < x ≤ 1 .

(b) The easy example to write down is Dirichlet’s function on [0 , 1]: f ( x ) =

(

1 q if x = p q in lowest terms

0 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 x = c with 0 < c < 1 by the intermediate value theorem. But when x = c , it follows that 2 x

= 3 x .

1