Log in
HomeStudy Guides420,000US230,000

MATH 140 Midterm: MATH140H BOYLE-M FALL2004 0101 MID SOL

36 views2 pages
turquoisegnu128
15 Feb 2019
School
University of Maryland
Department
Mathematics
Course
MATH 140
Professor
All

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

No books, no notes, no calculators: (a) (5 points) suppose a and l are real numbers and f is a function de ned on an open interval containing a. Let = min{1, /8}; that is, is the smaller of the two numbers. 1 and /8. (note that |x 3| < 1 means 2 < x < 4. ) 0 < |x 3| < = |(x2 + x) 12| < |4 + 4| (because 2 < x < 4, we have |x + 4| < 8) < 8 8( /8) = . (30 points) evaluate each of the following. 2. number, + , , or does not exist . No proof required. (a) lim x 0 (b) lim x e sin(5x) sin(7x) ln(ln(x)) ln(x)) 7 (c) lim x /2+ sec(x) = + (d) lim x 0 x sin(1/x) = 0 . (for all x, |x| sin(1/x) |x|.

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers

Related textbook solutions

Calculus
4 Edition,
Rogawski
ISBN: 9781319050733
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
CALCULUS:EARLY TRANSCENDENTALS
4 Edition,
Rogawski
ISBN: 9781319050740

Related Documents

MATH 140
Final Exam
Study Guide
MATH 140 Midterm: MATH140H BOYLE-M FALL2010 0101 MID EXAM
turquoisegnu128
MATH 140 Lecture Notes - Lecture 3: Quotient Rule, Product Rule
cerisegiraffe538
MATH 140
Final Exam
Study Guide
MATH 140 Midterm: MATH140H BOYLE-M FALL1997 0101 MID EXAM 3
turquoisegnu128

Related Questions

10. Give an e-d proof that lim(-2x + 1) = -3. Solution: Pick any e > 0 and let 8=. If |x - 21 < 8, then f (x) - L = 1 - 2.0 +1-(-3) = 1 - 2.0 + 4 = -2|- 21

[3] 7. Give an e-8 proof that lim (1 – 3.r) = 4. Solution: Pick any e > 0 and let := If +11 <0, then \f(x) - L = |(1 – 3x) - 4 = 1 - 3x - 31 = 1 - 3||* +11

4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.