This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Problem Set 1 Solutions Economics 6170 TA: Christopher Handy September 15, 2008 Problem 1 Applying the definition of continuity, we can say that f ( x ) = 1 1+ x is continuous on R + if for any choice of x 0, we have lim x x f ( x ) = f ( x ). Appealing now to the definition of a limit, we want to show that given any > 0, there is some ( ) > 0 such that if x 0 and | x- x | < , then | f ( x )- f ( x ) | < . To that end, fix an arbitrary x 0 and > 0. Lets look at the term we want be smaller than : for any x 0, we have | f ( x )- f ( x ) | = 1 1 + x- 1 1 + x = x- x (1 + x )(1 + x ) = | x- x | (1 + x )(1 + x ) . Now, since 1 + x 1 and 1 + x 1, we have that 0 < 1 (1+ x )(1+ x ) 1. That means we can simply take ( ) = : whenever x 0 satisfies | x- x | < , we have | f ( x )- f ( x ) | = | x- x | (1 + x )(1 + x ) < = . This shows f ( x ) = 1 1+ x is continuous on R + . Problem 2 First, note that we can write f ( x ) = e x ln a because e x ln a = [ e ln a ] x = a x . This suggests choosing g ( x ) = e x and h ( x ) = x ln a so that f ( x ) = g ( h ( x )). Since f : R R , g : R R , and h : R R , the chain rule result applies. It tells us that, since, the chain rule result applies....
View Full Document
- Fall '08