{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–3. Sign up to view the full content.

Tran, Tony – Homework 3 – Due: Sep 11 2007, 3:00 am – Inst: Samuels 1 This print-out should have 16 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 3) 10 points (i) Determine the value of lim x 5+ x - 6 x - 5 . 1. limit = -∞ correct 2. limit = 3. limit = 6 5 4. limit = - 6 5 5. none of the other answers Explanation: For 5 < x < 6 we see that x - 6 x - 5 < 0 . On the other hand, lim x 5+ x - 5 = 0 . Thus, by properties of limits, lim x 5+ x - 6 x - 5 = -∞ . 002 (part 2 of 3) 10 points (ii) Determine the value of lim x 5 - x - 6 x - 5 . 1. limit = 6 5 2. limit = correct 3. limit = -∞ 4. none of the other answers 5. limit = - 6 5 Explanation: For x < 5 < 6 we see that x - 6 x - 5 > 0 . On the other hand, lim x 5 - x - 5 = 0 . Thus, by properties of limits, lim x 5 - x - 6 x - 5 = . 003 (part 3 of 3) 10 points (iii) Determine the value of lim x 5 x - 6 x - 5 . 1. limit = -∞ 2. none of the other answers correct 3. limit = - 6 5 4. limit = 5. limit = 6 5 Explanation: If lim x 5 x - 6 x - 5 exists, then lim x 5+ x - 6 x - 5 = lim x 5 - x - 6 x - 5 . But as parts (i) and (ii) show, lim x 5+ x - 6 x - 5 6 = lim x 5 - x - 6 x - 5 . Consequently, lim x 5 x - 6 x - 5 does not exist .

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

View Full Document
Tran, Tony – Homework 3 – Due: Sep 11 2007, 3:00 am – Inst: Samuels 2 keywords: limit, left hand limit, right hand limit, rational function, 004 (part 1 of 1) 10 points Suppose that f ( x ) is defined for all x in U = (1 , 2) (2 , 3) and that lim x 2 f ( x ) = L. Which of the following statements is then true? I) If L > 0, then f ( x ) > 0 on U . II) If f ( x ) > 0 on U , then L 0. III) If L = 0, then f ( x ) = 0 on U . 1. I, II only 2. II only correct 3. II, III only 4. each of I, II, III 5. None of these 6. I, III only Explanation: I) False: consider the function f ( x ) = 1 - 2 | x - 2 | . Its graph is 2 4 6 so lim x 2 f ( x ) = 1 . But on (1 , 3 2 ) and on ( 5 2 , 3) we see that f ( x ) < 0. II) True: if f ( x ) > 0 on U , then on U the graph of f always lies above the x -axis. So as x approaches 2, the point ( x, f ( x )) on the graph approaches the point (2 , L ). Thus L 0; notice that L = 0 can occur as the graph 2 4 6 2 of f ( x ) = | x - 2 | shows.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online