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18.100pset9

# 18.100pset9 - 18.100B/C Fall 2008 Solutions to Homework 9 1...

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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 Taylor’s Theorem for the degree-3 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 0 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

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for all polynomials Q of degree n . Let P be a polynomial of degree n + 1. Then as x → ∞ , P ( x ) e x is of the form ±∞ and so we can apply l’Hospital’s rule to get lim x →∞ P ( x ) e x
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18.100pset9 - 18.100B/C Fall 2008 Solutions to Homework 9 1...

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