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Problem_Set_1_Solutions

# Problem_Set_1_Solutions - Problem Set 1 Solutions Economics...

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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 ε > 0, there is some δ ( ε ) > 0 such that if x 0 and | x - ¯ x | < δ , then | f ( x ) - f x ) | < ε . To that end, fix an arbitrary ¯ x 0 and ε > 0. Let’s look at the term we want be smaller than ε : 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 δ ( ε ) = ε : whenever x 0 satisfies | x - ¯ x | < δ , we have | f ( x ) - f x ) | = | ¯ x - x | (1 + x )(1 + ¯ x ) < δ = ε. 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 g 0 ( x ) = e x and h 0 ( x ) = ln a , we have f 0 ( x ) = g 0 ( h ( x )) h 0 ( x ) = e x ln a ln a = a x ln a.

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