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

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Unformatted text preview: MATH 137 Solutions Assignment 9 Part 2 Was due November 25 2011 1. We know that 1 e x → 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 ≥ . 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 ‘ = lim x →∞ x e x = lim x →∞ e- x = 0 , hence P (0) is true . Assume that for some n ≥ , 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 ) ( e x ) ( 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 ≥ 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 ≥ , as desired. 2. The function f ( x ) = sin x x for x 6 = 0 1 for x = 0 is continuous at . This is because sin x x → 1 as x → . By putting f (0) = 1 we have removed the potential discontinuity at . Show that f is differentiable at , and find f (0) . Hint. Look at the Newton quotient for f at and use L’Hospital’s rule. Solution: Recall that f is differentiable at if f (0) = lim x → f ( x )- f (0) x = lim x → sin x x- 1 x 1 exists. Since we know that lim x → sin x x = 1 , the limit form of f (0) is clearly indeterminate form of the type . Hence we can use l’Hospital’s rule and we see that f (0) = lim x → ( sin x x- 1) x provided this limit exist. But this limit equals lim x → x cos x- sin x x 2 1 = lim x → x cos x- sin x x 2 . This limit is again indeterminate of the form , so by l’Hospital’s rule it exists, provided lim x → ( x cos x- sin x ) ( x 2 ) = lim x → cos x- x sin x- cos x 2 x = lim x →- sin x 2 exists. It is clear that this last limit exists and is 0, therefore f (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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This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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

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