advertisement

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 δ = M.

(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