hw 1 - rhodes(ajr2283 – HW01 – Gilbert –(56380 1 This...

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Unformatted text preview: rhodes (ajr2283) – HW01 – Gilbert – (56380) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Welcome to Quest. Print off this as- signment and bring it to lectures as well as discussion groups. 001 10.0 points Determine if lim x →− 1 parenleftBig 4 x 2 + 3 x 2 + 1 parenrightBig exists, and if it does, find its value. 1. limit does not exist 2. limit = 7 2 correct 3. limit = 3 4. limit = 4 5. limit = 7 Explanation: Set f ( x ) = 4 x 2 + 3 , g ( x ) = x 2 + 1 . Then lim x →− 1 f ( x ) = 7 , lim x →− 1 g ( x ) = 2 . Thus the limits for both the numerator and denominator exist and neither is zero; so L’Hospital’s rule does not apply. In fact, all we have to do is use properties of limits. For then we see that limit = 7 2 . 002 10.0 points Determine the value of lim x →∞ x √ x 2 + 6 . 1. limit = 1 4 2. limit = 1 2 3. limit = ∞ 4. limit = 1 correct 5. limit = 2 6. limit = 0 7. limit = 4 Explanation: Since lim x →∞ x √ x 2 + 6 → ∞ ∞ , the limit is of indeterminate form. We might first try to use L’Hospital’s Rule lim x →∞ f ( x ) g ( x ) = lim x →∞ f ′ ( x ) g ′ ( x ) with f ( x ) = x , g ( x ) = radicalbig x 2 + 6 to evaluate the limit. But f ′ ( x ) = 1 , g ′ ( x ) = x √ x 2 + 6 , so lim x →∞ f ′ ( x ) g ′ ( x ) = lim x →∞ √ x 2 + 6 x = ∞ ∞ , which is again of indeterminate form. Let’s try using L’Hospital’s Rule again but now with f ( x ) = radicalbig x 2 + 6 , g ( x ) = x , and f ′ ( x ) = x √ x 2 + 6 , g ′ ( x ) = 1 . In this case, lim x →∞ √ x 2 + 6 x = lim x →∞ x √ x 2 + 6 , rhodes (ajr2283) – HW01 – Gilbert – (56380) 2 which is the limit we started with. So, this is an example where L’Hospital’s Rule applies, but doesn’t work! We have to go back to algebraic methods: x √ x 2 + 6 = x | x | radicalbig 1 + 6 /x 2 = 1 radicalbig 1 + 6 /x 2 for x > 0. Thus lim x →∞ x √ x 2 + 6 = lim x →∞ 1 radicalbig 1 + 6 /x 2 , and so limit = 1 . 003 10.0 points Determine if the limit lim x → 1 x 4- 1 x 3- 1 exists, and if it does, find its value. 1. limit = 4 3 correct 2. limit =-∞ 3. none of the other answers 4. limit = ∞ 5. limit = 3 4 Explanation: Set f ( x ) = x 4- 1 , g ( x ) = x 3- 1 . Then lim x → 1 f ( x ) = 0 , lim x → 1 g ( x ) = 0 , so L’Hospital’s rule applies. Thus lim x → 1 f ( x ) g ( x ) = lim x → 1 f ′ ( x ) g ′ ( x ) . But f ′ ( x ) = 4 x 3 , g ′ ( x ) = 3 x 2 . Consequently, limit = 4 3 . 004 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 g ( x ) G ( x ) ; C. lim x → 1 f ( x ) g ( x ) ; are indeterminate forms?...
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hw 1 - rhodes(ajr2283 – HW01 – Gilbert –(56380 1 This...

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