Exam3 8 - are diFerentiable functions such that lim x f ( x...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 8 2. limit = 2 3 correct 3. limit = 10 17 4. limit = 5 8 5. limit = 4 5 Explanation: Set f ( x ) = 3 tan 3 x + x and g ( x ) = 4 tan 5 x - 5 x. Then, except at odd multiples of π 2 , these functions are everywhere diFerentiable; in ad- dition, lim x 0 f ( x ) = 0 , lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule applies: lim x 0 f ( x ) g ( x ) = lim x 0 f 0 ( x ) g 0 ( x ) . Now f 0 ( x ) = 9 sec 2 3 x + 1 , while g 0 ( x ) = 20 sec 2 5 x - 5 , so lim x 0 f 0 ( x ) = 10 , lim x 0 g 0 ( x ) = 15 . Consequently, lim x 0 3 tan 3 x + x 4 tan 5 x - 5 x = 2 3 . keywords: 014 (part 1 of 1) 10 points Determine lim x 0 e 2 x - 2 x - 1 9 x 2 · . 1. limit doesn’t exist 2. limit = 11 9 3. limit = - 5 18 4. limit = 13 18 5. limit = 2 9 correct Explanation: The limit in question is of the form: lim x 0 f ( x ) g ( x ) where f and g
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Unformatted text preview: are diFerentiable functions such that lim x f ( x ) = 0 , lim x g ( x ) = 0 . Thus LHospitals rule can be applied, so lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . But lim x f ( x ) = 0 , lim x g ( x ) = 0 , in which case LHospitals rule has to be ap-plied again. Consequently, lim x f ( x ) g ( x ) = lim x f 00 ( x ) g 00 ( x ) . Since f 00 ( x ) = 4 e 2 x , g 00 ( x ) = 18 , it now follows that lim x e 2 x-2 x-1 9 x 2 = 2 9 ....
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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