MATH 231 Study Guide - Quiz Guide: Asymptote

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aditya

29 May 2020

School

UNC

Department

Mathematics

Course

MATH 231

Professor

Korkeat Korkeathikhun

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Document Summary

Due: tuesday, may 26: (limits at in nity with a radical. ) 1 x3 = 5. x3 from the denominator: Let"s compute the limit on the left that is going to . We factor out x6 from the numerator and. 3 1(cid:1) where we split the square root in the numerator using the general rule: a b = a b. X2 = |x| |x| |x| = |x|3. X2 fact that x2 = |x| to conclude that x6 = x2 x2 x2 = x2 x qx6(cid:0)5 + 2. Because we are taking x , we make the assumption that all x-values are positive, so we can take |x| = x. Making this replacement in the numerator of ( ) gives. 5 x x3 q5 + 2 x3(cid:0) 1. 5. after cancelling the x3 terms from the numerator and denominator. 2/x5 0 and 1/x3 0 as x goes to , we have that the limit is.

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.

What are the answers for all of these questions? I am confused on how to solve for these.