FINAL REVIEW #1-solutions

# FINAL REVIEW #1-solutions - suleimenov(bs26835 – FINAL...

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Unformatted text preview: suleimenov (bs26835) – FINAL REVIEW #1 – rusin – (55565) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Here is a review for the final exam. I am posting this in several parts. 001 10.0 points When f, g, F and G are functions such that lim x → 1 f ( x ) = 0 , lim x → 1 g ( x ) = 0 , lim x → 1 F ( x ) = 2 , lim x → 1 G ( x ) = ∞ , which, if any, of A. lim x → 1 f ( x ) g ( x ) , B. lim x → 1 F ( x ) g ( x ) , C. lim x → 1 g ( x ) G ( x ) , are indeterminate forms? 1. B and C only 2. A only 3. A and B only 4. B only 5. all of them 6. none of them correct 7. A and C only 8. C only Explanation: A. By properties of limits lim x → 1 f ( x ) g ( x ) = 0 · 0 = 0 , so this limit is not an indeterminate form. B. By properties of limits lim x → 1 F ( x ) g ( x ) = 2 = 1 , so this limit is not an indeterminate form. C. By properties of limits lim x → 1 g ( x ) G ( x ) = ∞ = 0 , so this limit is not an indeterminate form. 002 10.0 points Determine if lim x →∞ x 6 parenleftBig e 3 /x − 1 parenrightBig exists, and if it does, find its value. 1. limit = 2 2. limit = 1 2 correct 3. limit = 6 4. limit does not exist 5. limit = 0 6. limit = 3 Explanation: Since the limit has the form ∞ × 0, it is indeterminate and L’Hospital’s Rule can be applied: lim x →∞ f ( x ) g ( x ) = lim x →∞ f ′ ( x ) g ′ ( x ) with f ( x ) = e 3 /x − 1 , g ( x ) = 6 x , and f ′ ( x ) = − 3 x 2 e 3 /x , g ′ ( x ) = − 6 x 2 . In this case, lim x →∞ f ( x ) g ( x ) = lim x →∞ 1 2 e 3 /x . suleimenov (bs26835) – FINAL REVIEW #1 – rusin – (55565) 2 On the other hand, setting s = 1 /x we see that lim x →∞ e 3 /x = lim s → 0+ e 3 s = 1 . Consequently, the limit exists and limit = 1 2 . 003 10.0 points Determine if the improper integral I = integraldisplay ∞ 4 (3 x + 2) 2 dx is convergent, and if it is, find its value. 1. I = 1 2. I = 1 3 3. I is not convergent 4. I = 2 3 correct 5. I = 1 2 6. I = 5 6 Explanation: The integral is improper because the in- terval of integration is infinite. To test for convergence we thus have to determine if lim t →∞ I t , I t = integraldisplay t 4 (3 x + 2) 2 dx , exists. To evaluate I t , set u = 3 x + 2, Then du = 3 dx , while x = 0 = ⇒ u = 2 x = t = ⇒ u = 3 t + 2 . In this case, I t = 1 3 integraldisplay 3 t +2 2 4 u 2 du = 1 3 bracketleftBig − 4 u bracketrightBig 3 t +2 2 , so that lim t →∞ I t = lim t →∞ parenleftBig 2 3 − 4 3(3 t + 2) parenrightBig = 2 3 . Consequently, the integral is convergent and I = 2 3 . . 004 10.0 points Compute the volume of the solid of revolu- tion obtained by rotating the region R = braceleftBig ( x, y ) : x ≥ 3 , ≤ y ≤ 4 x bracerightBig about the x-axis....
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FINAL REVIEW #1-solutions - suleimenov(bs26835 – FINAL...

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