111FinalSpring2008-Solutions

111FinalSpring2008-Solutions - MATH 111, Spring 2009 Final...

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Unformatted text preview: MATH 111, Spring 2009 Final Exam Solutions Question 1. (a) lim x →∞ x 3 + 2 x + 7 ( x 2- 1)(2 x + 3) = lim x →∞ x 3 + 2 x + 7 2 x 3 + 3 x 2- 2 x- 3 = lim x →∞ 1 x 3 · ( x 3 + 2 x + 7) 1 x 3 · (2 x 3 + 3 x 2- 2 x- 3) = lim x →∞ 1 + 2 x 2 + 7 x 3 2 + 3 x- 2 x 2- 3 x 3 = 1 + 0 + 0 2 + 0-- = 1 / 2 . (b) Since sin x → 0 as x → 0, this limit is indeterminate of the form [0 / 0]. Applying L’Hˆopital’s Rule , we calculate lim x → sin x x = lim x → d dx sin x d dx x = lim x → cos x 1 = cos(0) = 1 . Note that we could evaluate the last limit the way we did because the cosine function is continuous. (c) Note that- 1 ≤ sin x ≤ 1 for all real numbers x . In considering the limit as x → ∞ , we can assume that x > 0. For any x > 0, dividing each part of the above inequalities by x doesn’t change the direction of the inequalities, and we obtain- 1 x ≤ sin x x ≤ 1 x . As x → ∞ , we have- 1 x → 0 and 1 x → 0. By the Squeeze Theorem , this implies that lim x →∞ sin x x = . (d) As x → ∞ , we have 1 x → 0 and, since the sine function is continuous, we know sin ( 1 x ) → sin(0) = 0. Therefore the limit under consideration in this problem is indeterminate of type [ ∞· 0]. We can rewrite it as lim x →∞ x sin(1 /x ) = lim x →∞ sin(1 /x ) 1 /x , 1 MATH 111, Spring 2009 Final Exam Solutions thus obtaining an indeterminate form of type [0 / 0] to which we can apply L’Hˆopital’s Rule , yielding lim x →∞ x sin(1 /x ) = lim x →∞ sin(1 /x ) 1 /x ) = lim x →∞ d dx sin(1 /x ) d dx 1 /x ) = lim x →∞- cos(1 /x ) /x 2- 1 /x 2 = lim x →∞ cos(1 /x ) . As x → ∞ we have 1 x → 0, and since cosine is a continuous function, we conclude that the above limit is equal to cos(0) = 1 . (e) Recall that the derivative of a function f at the point a is f ( a ) = lim h → f ( a + h )- f ( a ) h . Therefore, the given limit is equal to d dx sin x π/ 3 = (cos x ) | π/ 3 = cos( π/ 3) = 1 / 2 . Question 2. (a) f ( x ) = d dx ln( x tan- 1 x ) = 1 x tan- 1 x · d dx x tan- 1 x = 1 x tan- 1 x · d dx x · tan- 1 x + x · d dx tan- 1 x = 1 x tan- 1 x · tan- 1 x + x 1 + x 2 . (b) By the Fundamental Theorem of Calculus , we have f ( x ) = d dx x 1 2 t dt = 2 x . (c) I don’t know why there are two problems listed here, but I will do both. Implicitly differenti- ating the first one, we obtain 3 x 2 + 3 y 2 dy dx- 9 ( y + x dy dx ) = 0. Plugging in the point ( x, y ) = (2 , 4) 2 MATH 111, Spring 2009 Final Exam Solutions yields 3 · 2 2 + 3 · 4 2 · dy dx (2 , 4)- 9 4 + 2 · dy dx (2 , 4) = 0 , 12 + 48 · dy dx (2 , 4)- 36- 18 · dy dx (2 , 4) = 0 , 30 · dy dx (2 , 4) = 24 , dy dx (2 , 4) = 24 / 30 = 4 / 5 ....
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This note was uploaded on 03/01/2012 for the course MATH 1110 at Cornell.

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111FinalSpring2008-Solutions - MATH 111, Spring 2009 Final...

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