HW03-solutions

# HW03-solutions - vasquez(mpv244 – HW03 – Schultz...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: vasquez (mpv244) – HW03 – Schultz – (56445) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Below is the graph of a function f . 2 4 − 2 − 4 2 4 − 2 − 4 Use the graph to determine lim x → 2 f ( x ). 1. limit = − 3 correct 2. limit = − 2 3. limit = − 1 4. limit = 0 5. does not exist Explanation: From the graph it is clear that the limit lim x → 2 − f ( x ) = − 3 , from the left and the limit lim x → 2+ f ( x ) = − 3 , from the right exist and coincide in value. Thus the two-sided lim x → 2 f ( x ) = − 3 . 002 10.0 points Below is the graph of a function y = f ( x ) 2 4 6 − 2 − 4 − 6 2 4 6 8 − 2 − 4 Use the graph to determine lim x →− 1+ f ( x ) . 1. limit = − 4 2. limit = 1 correct 3. limit = 5 2 4. limit does not exist 5. limit = 4 Explanation: From the graph we see that lim x →− 1+ f ( x ) = 1 . 003 10.0 points If f oscillates faster and faster when x ap- proaches 0 as indicated by its graph vasquez (mpv244) – HW03 – Schultz – (56445) 2 determine which, if any, of L 1 : lim x → 0+ f ( x ) , L 2 : lim x → − f ( x ) exist. 1. neither L 1 nor L 2 exists 2. both L 1 and L 2 exist 3. L 1 doesn’t exist, but L 2 does correct 4. L 1 exists, but L 2 doesn’t Explanation: For x > 0 the graph of f oscillates but the oscillations do not get smaller and smaller as x approaches 0; so L 1 does not exist. But for x < 0, the graph oscillates and the oscillations get smaller and smaller as x approaches 0; in fact, the oscillation goes to 0 as x approaches 0, so L 2 exists. Consequently, L 1 does not exist, but L 2 does . 004 10.0 points Consider the function f ( x ) = 1 − x, x < − 1 x, − 1 ≤ x < 2 ( x − 1) 2 , x ≥ 2 . Find all the values of a for which the limit lim x → a f ( x ) exists, expressing your answer in interval no- tation. 1. ( −∞ , − 1] ∪ [2 , ∞ ) 2. ( −∞ , ∞ ) 3. ( −∞ , 2) ∪ (2 , ∞ ) 4. ( −∞ , − 1) ∪ ( − 1 , ∞ ) 5. ( −∞ , − 1) ∪ ( − 1 , 2) ∪ (2 , ∞ ) correct Explanation: The graph of f is a straight line on ( −∞ , − 1), so lim x → a f ( x ) exists (and = f ( a )) for all a in ( −∞ , − 1). Similarly, the graph of f on ( − 1 , 2) is a straight line, so lim x → a f ( x ) exists (and = f ( a )) for all a in ( − 1 , 2). On (2 , ∞ ), however, the graph of f is a parabola, so lim x → a f ( x ) still exists (and = f ( a )) for all a in (2 , ∞ ). On the other hand, lim x →− 1 − f ( x ) = 2 , lim x →− 1+ f ( x ) = − 1 , while lim x → 2 − f ( x ) = 2 , lim x → 2+ f ( x ) = 1 ....
View Full Document

## This note was uploaded on 04/18/2009 for the course CH 52375 taught by Professor Ruth during the Spring '09 term at University of Texas.

### Page1 / 11

HW03-solutions - vasquez(mpv244 – HW03 – Schultz...

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

View Full Document
Ask a homework question - tutors are online