This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 nonnegative 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 ) ....
View
Full
Document
This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Algebra, Mathematical Induction

Click to edit the document details