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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 9 1. Let f ( x ) = sin x . Then f (0) = 0, f (0) = cos 0 = 1 and f (0) = sin 0 = 0. Using Taylors Theorem for the degree3 expansion for f ( x ) around the point x = 0, we have for any x R that f ( x ) = x 1! + f (3) ( t ) 3! x 3 for some t between 0 and x . We have f (3) ( x ) = cos x , so we can rewrite this equation as f ( x ) x = cos t 6 x 3 . Taking absolute values and using the fact that  cos t  1 for all t R gives  sin( x ) x   x  3 6 , as needed. For the second part of the question, we always consider x and so also x > 0. From above we have  sin( x c ) x c  1 6 x 3 c and so  x a sin( x c ) x a c  1 6 x a 3 c . If a < c then x a c 0 and so  x a sin( x c )  has the same limit as  x a sin( x c ) x a c  . This latter expression is bounded between 0 and 1 6 x a 3 c 0, so in fact in this case x a sin x c 0. If a = c then we have  x a sin( x c ) 1  1 6 x 2 a and so x a sin x c 1. Finally, if a > c then x a sin x c = x a c ( x c sin x c ). Since x a c and x c sin x c 1 we have x a sin x c . 2. For the first part of the question, we proceed by induction on the degree of the polynomial. We know that lim x e x = and so lim x c e x = 0 for any c R . Suppose that lim x Q ( x ) e x = 0 1 for all polynomials Q of degree n . Let P be a polynomial of degree n + 1. Then as x...
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.
 Fall '10
 Prof.KatrinWehrheim

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