HW5Solution

# HW5Solution - Sketch of solution of HW 5 1 2 r b=0 So you...

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Sketch of solution of HW 5 1. r 2 b = 0 So, you need to consider three cases : b < 0, b = 0 and b > 0. b < 0 : y 1 = x b , y 2 = x − b So, the limit corresponds to y 2 doesn't exist. b = 0 : y 1 = 1 , y 2 = ln x So, the limit corresponds to y 2 doesn't exist. b > 0 : y 1 = cos  b ln x , y 2 = sin b ln x So, the limit doesn't exist for both of them. 2. Done in class. 3. x 1 = u , x 2 = u' , x 3 = u ' ' Then x 3 ' =− x 1 . 4. The inverse is 1 14 ( 3 -4 1 ) 5 -2 -3 -7 14 7 5. Similar to Q10 in section 7.3. The answer is the same x 1 x 2 x 4 = 0 . 6. W = t 2 t 2 , so the interval will be everything except t = 0, 2 , − 2 . 7. r 1 = 2  3 ,r 2 =− 2, r 3 = 2 − 3 1 = 1 1 3 , 1 3 , 1 T , 2 =− 1,
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## This note was uploaded on 07/29/2008 for the course MATH 235 taught by Professor Chen during the Fall '08 term at Michigan State University.

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