Homework 19

Homework 19 - vu(tv2894 – Homework 19(Section 7.2 –...

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Unformatted text preview: vu (tv2894) – Homework 19 (Section 7.2) – miner – (55096) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x →∞ parenleftbigg e − 2 x- 5 e 2 x 4 e 2 x + e − 2 x parenrightbigg exists, and if it does, find its value. 1. limit = 1 4 2. limit = 1 3. limit =- 5 4. limit =- 5 4 correct 5. limit = 0 6. limit does not exist Explanation: Since lim x →∞ e − ax = 0 , lim x →∞ e ax = ∞ when a > 0, taking limits directly gives lim x →∞ parenleftbigg e − 2 x- 5 e 2 x 4 e 2 x + e − 2 x parenrightbigg = ∞ ∞ , which doesn’t make any sense. (And we can’t just cancel the ∞ ’s because infinities don’t work like that.) So we try to get rid of terms that go to ∞ and leave terms that go to zero. To achieve this, multiply top and bottom by e − 2 x . Then e − 2 x e − 2 x parenleftbigg e − 2 x- 5 e 2 x 4 e 2 x + e − 2 x parenrightbigg = e − 4 x- 5 4 + e − 4 x . In this case lim x →∞ e − 2 x- 5 e 2 x 4 e 2 x + e − 2 x = lim x →∞ e − 4 x- 5 4 + e − 4 x . But, as we noted, lim x →∞ e − 4 x = 0 . so by properties of limits, lim x →∞ ( e − 4 x- 5) =- 5 , lim x →∞ (4 + e − 4 x ) = 4 . Thus by properties of limits yet again, lim x →∞ parenleftbigg e − 2 x- 5 e 2 x 4 e 2 x + e − 2 x parenrightbigg exists and the limit =- 5 4 ....
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This note was uploaded on 11/07/2010 for the course M 408N taught by Professor Gualdini during the Fall '10 term at University of Texas.

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Homework 19 - vu(tv2894 – Homework 19(Section 7.2 –...

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