MATH 1300 Midterm: MATH 1300 Test1Sol

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turquoisegnu128

31 Jan 2019

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York University

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Mathematics and Statistics

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MATH 1300

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All

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$MATH 1300 Full Course Notes$

MATH 1300 Full Course Notes

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1 + x 1 x x (10 pts) 1 + x 1 x x lim x 0. = lim x 0 (1 + x) (1 x) x( 1 + x + 1 x) = 1: find the following limits using the de nition of limit. (a) lim x 0. For any > 0, let = . Without loss of generality we can assume < 1, then for all. |x| < , we have |2x + 4| > 1. = |x| f (x) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 2 (b) lim x x2 + 1 x2 + 2 (10 pts) For any > 0, let n = , then for all x > n , we have. |f (x) 1| = (cid:12) (cid:12) (cid:12) (cid:12) x2 + 1 x2 + 2 1 (cid:12) (cid:12) (cid:12) (cid:12)

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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.

4. (21 pts) (I) (6 pts) Determine the value of a constant b for which the function g is continuous at 0. Show your work. Justify your answer! 2 x +b if x<0, X-5 g(x) = if x 20, x#4. We have to show that lim g (x) = g (0). g(0) = 0 +18 = -1; lim X0* 0 + 16 ) = lim x + 16 **0* x2 - 160 - 16- Since lim g (x) has to exist (in order for g to be continuous at x = 0), it has to be true that x-0- x0+ X- 0 lim g(x) - lim g(x) = lim g(x). Therefore, -1, and , so b = 5. so, if b = 5, lim g (x) = g(0), and therefore, g is continuous at x = 0. (II) (4 pts) Evaluate the limit. Show your work! 2. b limx- (a) lim g(x) = lim *+- 2 x + b X - 5 : 1 * +16 10 g (x) = lim x++ - Elim x2 - 16 x++ (x - 10). OR 2x+5. 2; (a) lim g (x) = lim x--00 X- (b) lim x++ 2 + 16 g(x) = lim **** x2 - 16 X-+ (III) (4 pts) Determine whether g has any horizontal asymptotes. If so, write the equation (or equations) of all horizontal asymptotes. y = 2, by part (II) (a) y = 0, by part (II) (b) (IV) (7 pts) Determine whether g has any vertical asymptotes. If so, write the equation (or equations) of all vertical asymptotes. Show your work. Justify your answer! 5 + 16 There is no vertical asymptote at x = 5, because lim g(x) = x+5 52-16' -8 + b (x) = there is no vertical asymptote at x = -4, because lim X -4 - 4 - 5 But, lim g(x) = lim + 16 x+4+ x2 - 16 lim_ (x + 16) -20 x+4+ (x - 4) 60 (x + 4) +8 (or the numerator goes to 20, and the denominator is positive and goes to o, as x + 4*) = +00 So, the function g has only one vertical asymptote, x = 4.

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

MATH 1300 Midterm: MATH 1300 Test1Sol

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