MATH 13100 Lecture Notes - Lecture 4: Formal Proof

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sangriamuskrat676

20 Jun 2018

School

University of Chicago

Department

Mathematics

Course

MATH 13100

Professor

Ronno Das

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MATH 131

Delta/epsilon proofs: part II

Epsilon/Delta Definition of limits

if for every , there exists a (depending on epsilon) such that (x)lim

x→cf=Lε > 0 δ > 0 f(x)

|−L|< ε

provided , that is, 0 < x

|−c|< δ .0 < x

|−c|< δ ⇒f(x)

|−L|< ε

Epsilon/Delta Definition of One-sided limits

: given , there is a . (x)lim

x→cf=Lε > 0 δ > 0 .0 < x

|−c|< δ ⇒f(x)

|−L|< ε

c<x<c+d

c-d<x<c → L-E<f(x)<L+E

One-sided limits, depending on the side

Right hand: means that (x)lim

x→c+f=L0 < x−c< δ ⇒f(x)

|−L|< ε

Left hand: means that (x)lim

x→c+f=L c − δ < x<c⇒f(x)

|−L|< ε

Helpful Diagram:

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Document Summary

0 < x| c| < (cid:303) f(x) L| < (cid:304) (cid:303) > 0 f (x) f (x) Epsilon/delta definition of one-sided limits lim x c c 0 (x) 0 < x c < (cid:303) f(x) L| < (cid:304) c (cid:303) < x < c f(x) In the following, k is a constant, n positive integer, f, g are functions such that lim x c f (x) and lim x c (x) g. Then: k = k lim x c x = c lim x c f(x) k. )(x) x c * g lim (f/g)(x) lim x c. [lim (x)][lim (x)] g f x c lim (x))/(lim (x) f x c x c g x c (x) 8. lim x c: lim x c g lim x c n n = ( (f(x)) lim (x)) x c.

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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.

Suppose that the first and second derivatives of a function y=f(x) are given by: f'(x) = xe^x and f"(x) = e^x (x+1). Which of the following statements is true? a) f'(-1) > 0 b) f(x) is increasing and opens up (0, infinity) c) There are no critical points and there is one inflection point. d) f(x) is increasing and opens down on (negative infinity, -1) e) There are is one critical point and there are two inflection points. f) f"(x) > 0 on (negative infinity, -1) g) There are no critical points and there are two inflection points. h) f'(x) > 0 on (negative infinity, 0) i) f'(x) < 0 on (0, infinity) j) f"(x) < 0 on (-1, infinity)

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

MATH 13100 Lecture Notes - Lecture 4: Formal Proof

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