MATH 1131Q Lecture Notes - Lecture 2: Horse Length, Classification Of Discontinuities, Intermediate Value Theorem

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MATH 1131Q Lecture 1: new doc 2018-09-08 19.21.16_20180908192354

MATH 1131Q Lecture Notes - Lecture 2: Horse Length, Classification Of Discontinuities, Intermediate Value Theorem

MATH 1131Q Lecture Notes - Lecture 3: Exponential Function, Power Rule, Differentiable Function

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Math 1131q , lecture 2 , section 2. 4-2. 6. Example : is read as or x 1. The limit as x approaches a of f(x) is . The limit of f(x) a x approaches a lim x^2 = 1 and lim x^2 = 4 x 2. Read as , the limit of x^2 is 4 as x approaches 2 . If for every number (cid:304) > 0 there is a number (cid:303) > 0 such that. If 0 < i x - a i < (cid:303) then i f(x) - i < (cid:304) Definition - a function f is continuous at a number a if. For a function to be implicit it must be continuous at a and : lim f(x) = f(a) x a f(a) is defined. If there is a hole, it means that the value is not defined at the domain = discontinuous.

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1. (18 pts) The graph of a function f, defined on (-0, +o), is given in the figure below. - - y = f(x) - - - - - - N - - 4 A -2-1 144 -1 - - - - - - - -- - - -- Use the graph of f to answer questions (a)-(f) below. (a) Determine the value or write "does not exist". i. f(-3) = -2 ii. f(0) = 1 (b) Determine the limit or write "does not exist" (DNE). Write "does not exist" only if a limit does not exist and is not too or -0. i. lim f(x) = 0 iv. lim f(x) = +00 X- 3 x2 ii. lim f(x) = -4 v. lim f(x) = -0 X- 0- x -00 iii. lim f(x) = DNE vi. lim f(x) = -2 x0 x++00 (C) MULTIPLE CHOICE! Circle ONLY one answer! Find a number a for which the following statement is true (the function f is shown above): STATEMENT: The function f is continuous at a, and lim f(x) = 0. i. a=-3 iii. a=0 V. a=3 ii. a = -2 iv. a = 2 vi. Such a number does not exist. (d) Determine the largest intervals of continuity of f. Use interval notation to write your answer. (-0, -3), (-3,0), (0, 2), (2, +00) (e) Write the equation(s) of all vertical asymptotes of f. Write "none" if appropriate. x= 2 (f) Write the equation(s) of all horizontal asymptotes of f. Write "none" if appropriate. y = -2

[ -2 e* if x< 0 3. (24 pts) Let g(x) = 1 X+6 if x20. [x - 3 (I) Evaluate the limit. (Note: Possible answers include + or - .) [You may not use a table of values, a graph, or L'Hospitals's Rule to justify your answer. ] SHOW YOUR WORK. (a) (3 pts) lim g(x) = ; (b) (3 pts) lim g(x) = x 10 x-10+ (c) (3 pts) lim g(x) = ; (d) (3 pts) lim g(x) = 14 = * 3+ (e) (3 pts) lim g(x) = ; (f) (3 pts) lim g(x) = X + (II) (6 pts) Using the definition of continuity, show that gis continuous at 0. SHOW YOUR WORK.

4. (21 pts) Let g be the function given by ( 6 - X 2 + x if x < 2, x + -2 g(x) = x2 – x-1 if x>2. (a) (6 pts) Use the definition of continuity to determine whether the function g is continuous at 2. Show your work. (b) (3 pts) Evaluate the limit. Justify your answer. Show your work. Write "does not exist" only if the limit does not exist and is not + or -0. lim g(x) = lim -2+ = +00 X 2 + X since the form of the limit is NUM: POS, DENOM: POS (c) (2 pts) Write the equation(s) of all vertical asymptotes of g. Write "none" if appropriate. The result in part (b) implies that the line x= -2 is a vertical asymptote. This is the only vertical asymptote, since the function g is continuous on its domain (d) (2 pts) Determine the largest) intervals of continuity of g. Use interval notation to write your answer. Let g be the function given by 12+ if x < 2, x# -2 2 + x g(x)= x2-x-1 x-1 if x>2. (e) (6 pts) Evaluate each limit. Do not use L'Hôpital's Rule. Write "does not exist" only if the limit does not exist and is not too or -00. Show your work. 6-1 i lim g(x) = lim -00 2 + X = lim -002 + x 0-1 + 1 x -00 x x x -+- 2 +1 0 0 + 11 x2 - x-1 1 ii. lim g(x) = lim x++ V x2 – x-1 X-1 x2 - x-1 X - 1 = lim x +oc x +0 - lim Vam M93001 (f) (2 pts) Write the equation(s) of all horizontal asymptotes of g. Write "none" if appropriate.

MATH 1131Q Lecture Notes - Lecture 2: Horse Length, Classification Of Discontinuities, Intermediate Value Theorem

MATH 1131Q Lecture 1: new doc 2018-09-08 19.21.16_20180908192354

MATH 1131Q Lecture Notes - Lecture 2: Horse Length, Classification Of Discontinuities, Intermediate Value Theorem

MATH 1131Q Lecture Notes - Lecture 3: Exponential Function, Power Rule, Differentiable Function

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