CAS MA 123 Lecture Notes - Lecture 5: Asymptote

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cerisegiraffe538

27 Sep 2015

School

Boston University

Department

Mathematics & Statistics

Course

CAS MA 123

Professor

John Bergdall

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Document Summary

Ma123- lecture 5 - techniques of calculating limits continued. Warm-up: lim x 1 x=1, lim x 1+ x x+1 is definitely . It has a vertical asymptote at x x +1 = : to determine this value look at x= - x x+1 = The lim x 1+ x x+1 = - : lim x 1- x x +1 = , to determine this value think a negative/a negative so. The lim x 1- x x +1 = + . Question: where are the vertical asymptotes of tan = sin cos . Sin 0 at these points > lim x . Graph lim x + cosx = dne same for x - . Use the squeeze theorem >>> -1 cosx 1. Definition: f(x) has a horizontal asymptote y=l where l is a finite number. If lim x f(x) then the limit = l.

Related Questions

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.

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.

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CAS MA 123 Lecture Notes - Lecture 5: Asymptote

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