{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 4 - Tran Tony Homework 4 Due 3:00 am Inst Samuels This...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Tran, Tony – Homework 4 – Due: Sep 19 2007, 3:00 am – Inst: Samuels 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the value of lim x 1 f ( x ) g ( x ) - 1 2 f ( x ) + 1 g ( x ) when lim x 1 f ( x ) = - 1 , lim x 1 g ( x ) = 4 . Correct answer: - 2 . 5 . Explanation: By properties of limits, lim x 1 f ( x ) g ( x ) - 1 2 f ( x ) + 1 g ( x ) = lim x 1 ( f ( x ) g ( x ) - 1) lim x 1 (2 f ( x ) + 1 g ( x )) . But again by properties of limits, lim x 1 ( f ( x ) g ( x ) - 1) = lim x 1 f ( x ) ·‡ lim x 1 g ( x ) · - 1 , while lim x 1 (2 f ( x ) + 1 g ( x )) = 2 lim x 1 f ( x ) · + 1 lim x 1 g ( x ) · . Consequently, limit = - 4 - 1 - 2 + 4 = - 5 2 ≈ - 2 . 5 . keywords: limit, laws of limits 002 (part 1 of 1) 10 points Below are the graphs of functions f and g . 4 8 - 4 4 8 - 4 - 8 f : g : Use these graphs to determine lim x 9 { f ( x ) + g ( x ) } . 1. limit = 7 2. limit = 0 3. limit = 3 4. limit = 4 correct 5. limit does not exist Explanation: From the graph it is clear that lim x 9 { f ( x ) + g ( x ) } = 4 . (Don’t forget that for a limit to exist at a point, the left and right hand limits have to exist and coincide. So determine left and right hand limits separately and use limit laws.) keywords: limit of sum of functions, graph, limit 003 (part 1 of 1) 10 points Determine lim x 2 n 1 x - 2 - 2 x 2 - 2 x o .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Tran, Tony – Homework 4 – Due: Sep 19 2007, 3:00 am – Inst: Samuels 2 1. limit = 2 2. limit = 1 2 correct 3. limit = - 1 3 4. limit = - 2 5. limit = - 1 2 6. limit = 1 3 7. limit does not exist Explanation: After simplification we see that 1 x - 2 - 2 x 2 - 2 x = x - 2 x ( x - 2) = 1 x , for all x 6 = 2. Thus limit = lim x 2 1 x = 1 2 . keywords: analytic limit, difference rational functions, limit, common denominators 004 (part 1 of 1) 10 points Determine if lim x 0+ 1 - (1 / x ) 1 + (4 / x ) exists, and if it does, find its value. 1. limit does not exist 2. limit = - 4 3. limit = - 1 4 correct 4. limit = 1 4 5. limit = 4 Explanation: After simplification and cancellation 1 - (1 / x ) 1+(4 / x ) = x - 1 x + 4 . On the other hand, lim x 0+ x = 0 , and so x - 1 x + 4 = - 1 4 by Properties of Limits. Consequently, the given limit exists and limit = - 1 4 . keywords: analytic limit, quotient radicals, keywords: 005 (part 1 of 1) 10 points Determine if lim x 1+ p 9 - x 2 exists, and if it does, find its value. 1. limit = 2 2 correct 2. limit = 11 3. limit = 1 4. limit does not exist 5. limit = 10 6. limit = 3 7. limit = 0 Explanation: For x near 1 the inequality 9 - x 2 > 0 holds, so f ( x ) = p 9 - x 2 is well defined for such x . Consequently, by Properties of Limits, the right hand limit lim x 1+ p 9 - x 2
Background image of page 2
Tran, Tony – Homework 4 – Due: Sep 19 2007, 3:00 am – Inst: Samuels 3 exists and limit = 2 2 . keywords: limit, one-sided limit, radical func- tion 006 (part 1 of 1) 10 points Determine if lim x 1 x 2 - 7 x + 6 x 2 + 4 x - 5 exists, and if it does, find its value.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}