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 Assignment No. 8 - Solutions Fall 2010 1. (a) lim x parenleftbigg cot x- 1 x parenrightbigg . This is an indeterminate form of the type - . If we put the two functions over a common denominator, i.e., lim x parenleftbigg cot x- 1 x parenrightbigg = lim x x cos x- sin x x sin x = lim x f ( x ) g ( x ) : a indeterminate form. If we now differentiate numerator and denominator, we obtain lim x f ( x ) g ( x ) = lim x - x sin x sin x + x cos x , another indeterminate form. Differentiating again (with faith), lim x f ( x ) g ( x ) = lim x - sin x- x cos x cos x + cos x- x sin x = 0. We can now work backward and conclude that lim x parenleftbigg cot x- 1 x parenrightbigg = lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) = 0. (b) lim x ( xe 1 /x- x ). This is an indeterminate form of the type - . If we factor the expression and then rewrite it as a product, i.e., ( xe 1 /x- x ) = x ( e 1 /x- 1) = e 1 /x- 1 1 /x , we have produced a indeterminate form since e 1 /x e = 1 as x . It is convenient to make the change of variable y = 1 x so that lim x e 1 /x- 1 1 /x = lim y e y- 1 y . Now differentiate numerator and denominator to find that lim y e y- 1 y = lim y e y 1 = 1. Therefore lim x ( xe 1 /x- x ) = 1. 2. Required to find a such that lim x parenleftbigg x + a x- a parenrightbigg x = e. (1) Method No. 1: Using the logarithm. If we take the natural logarithm of the LHS of (1), ln parenleftbigg x + a x- a parenrightbigg x = x ln parenleftbigg x + a x- a parenrightbigg = ln bracketleftbig ( x + a )( x- a ) 1 bracketrightbig x 1 . From (1), we would like the limit as x of this expression to be ln e = 1. In the limit x , the rightmost term is a indeterminate form. Rewrite this term as follows, f ( x ) g ( x ) = ln( x + a )- ln( x- a ) x 1 , 1 which implies that f ( x ) g ( x ) = ( x + a ) 1- ( x- a ) 1- x 2 = = 2 ax 2 ( x + a )( x- a ) . Then lim x f ( x ) g ( x ) = lim x 2 ax 2 ( x + a )( x- a ) = lim x 2 a (1 + ax 1 )(1- ax 1 ) = = 2 a. Recalling that this limit must be equal to ln e = 1, we have 2 a = 1, implying that a = 1 2 ....
View Full Document
This note was uploaded on 01/18/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.
- Fall '08