HW3 (solutions)

# HW3 (solutions) - myers(nmm698 – HW03 – Hamrick...

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Unformatted text preview: myers (nmm698) – HW03 – Hamrick – (54868) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Which of the following statements are true for all values of c ? I. lim x → c f ( x ) = 0 = ⇒ lim x → c | f ( x ) | = 0 . II. lim x → c | f ( x ) | = 0 = ⇒ lim x → c f ( x ) = 0 . 1. Both I and II correct 2. II only 3. Neither I nor II 4. I only Explanation: 002 (part 2 of 2) 10.0 points Which of the following statements are true for all c and all L ? I. lim x → c f ( x ) = L = ⇒ lim x → c | f ( x ) | = | L | . II. lim x → c | f ( x ) | = | L | = ⇒ lim x → c f ( x ) = L. 1. Both I and II 2. I only correct 3. II only 4. Neither I nor II Explanation: 003 10.0 points Below is the graph of a function f . 2 4 6 − 2 − 4 − 6 2 4 6 8 − 2 − 4 Use the graph to determine lim x → 3 f ( x ) . 1. limit = 18 2. limit = 9 3. limit does not exist correct 4. limit = 3 5. limit = 6 Explanation: From the graph it is clear the f has a left hand limit at x = 3 which is equal to 3; and a right hand limit which is equal to 0. Since the two numbers do not coincide, the limit does not exist . 004 10.0 points Determine the limit lim x → 7 1 ( x − 7) 2 . 1. limit = − 1 7 2. none of the other answers 3. limit = −∞ 4. limit = 1 7 myers (nmm698) – HW03 – Hamrick – (54868) 2 5. limit = ∞ correct Explanation: Since ( x − 7) 2 ≥ 0 for all x , we see that lim x → 7 1 ( x − 7) 2 = ∞ . keywords: limit, rational function 005 10.0 points By calculating the value of the function f ( x ) = (1 + x ) 1 /x successively at x = 0 . 01 , . 001 , . 0001 , . 00001 , . . . and x = − . 01 , − . 001 , − . 0001 , − . 00001 , . . . , estimate the value, E , of lim x → (1 + x ) 1 /x correct to 5 decimal places. 1. E ≈ 2 . 71826 2. E ≈ 2 . 71832 3. E ≈ 2 . 71828 correct 4. E ≈ 2 . 7183 5. E ≈ 2 . 71824 Explanation: By calculation we see that f (0 . 01) = 2 . 70481 , f (0 . 001) = 2 . 71692 , f (0 . 0001) = 2 . 71815 , f (0 . 00001) = 2 . 71827 . In the same way, we see that f ( − . 01) = 2 . 732 , f ( − . 001) = 2 . 71964 , f ( − . 0001) = 2 . 71842 , f ( − . 00001) = 2 . 7183 . This suggests that, to 5 decimal places, the value of f (0) lies between 2 . 71827 and 2 . 7183 . To determine exactly where it lies, we need to compute f ( x ) at the next smaller positive and negative values of x , i.e. , at x = 0 . 000001 , x = − . 000001 . Calculations show that f (0 . 000001) = 2 . 71828 , while f ( − . 000001) = 2 . 71828 . Consequently, to 5 decimal places, E ≈ 2 . 71828 . 006 10.0 points Determine the value of lim x → 3 parenleftBigg 3 g ( x ) − ( f ( x )) 2 3 f ( x ) + 4 parenrightBigg when lim x → 3 f ( x ) = − 1 , lim x → 3 g ( x ) = 3 ....
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HW3 (solutions) - myers(nmm698 – HW03 – Hamrick...

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