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soln9 - MATH 137 Solutions Assignment 9 Part 2 Was due 1 0...

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MATH 137 Solutions Assignment 9 Part 2 Was due November 25 2011 1. We know that 1 e x 0 as x → ∞ . Using the method of mathematical induction in conjunction with L’Hospital’s rule, show that lim x →∞ x n e x = 0 for every integer n 0 . Thus we learn that the exponential function grows faster than any integer power of x . Solution: Let n := lim x →∞ x n e x , and for n a non-negative integer, let P ( n ) be the statement “ n = 0 .” We compute 0 = lim x →∞ x 0 e x = lim x →∞ e - x = 0 , hence P (0) is true . Assume that for some n 0 , the statement P ( n ) is true. (That is our induction hypothesis). Now, n +1 = lim n →∞ x n +1 e x ( indeterminate form of type ) = lim n →∞ ( x n +1 ) 0 ( e x ) 0 ( by l’Hospital’s rule, provided the RHS exists ) = lim n →∞ ( n + 1) x n e x = ( n + 1) n = 0 ( since P ( n ) is assumed to be true ) . Hence we just proved that P ( n ) = P ( n + 1) for any n 0 By the principle of mathematical induction, P ( n ) is true for all n , that is lim x →∞ x n e x = 0 for every integer n 0 , as desired. 2. The function f ( x ) = sin x x for x 6 = 0 1 for x = 0 is continuous at 0 . This is because sin x x 1 as x 0 . By putting f (0) = 1 we have removed the potential discontinuity at 0 . Show that f is differentiable at 0 , and find f 0 (0) . Hint. Look at the Newton quotient for f at 0 and use L’Hospital’s rule. Solution: Recall that f is differentiable at 0 if f 0 (0) = lim x 0 f ( x ) - f (0) x = lim x 0 sin x x - 1 x 1
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exists. Since we know that lim x 0 sin x x = 1 , the limit form of f 0 (0) is clearly indeterminate form of the type 0 0 . Hence we can use l’Hospital’s rule and we see that f 0 (0) = lim x 0 ( sin x x - 1) 0 x 0 provided this limit exist. But this limit equals lim x 0 x cos x - sin x x 2 1 = lim x 0 x cos x - sin x x 2 . This limit is again indeterminate of the form 0 0 , so by l’Hospital’s rule it exists, provided lim x 0 ( x cos x - sin x ) 0 ( x 2 ) 0 = lim x 0 cos x - x sin x - cos x 2 x = lim x 0 - sin x 2 exists. It is clear that this last limit exists and is 0, therefore f 0 (0) = 0 . 3. In this exercise you will conduct a careful analysis of f ( x ) = x arctan x . (a) Decide if f is even, odd or neither. Solution: First note that domain( f ) = R . We have that f is even. Indeed, if y = x arctan x , we have x = tan y x . Since tan is odd ( sin is odd and cos is even), tan y - x = - tan y x = - x , so y - x = arctan( - x ) and thus y = ( - x ) arctan( - x ) .
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