2. Find the equation of the slant asymptote of the following function as 2 + +oo, if there is one: 2x -9 x2 + 5x x+5x 2x x Write your answer in the form y = mx + b. f(x) = 22 + x2 - 1

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5. (22 pts) if x < 2, Let g(x) = sin (1 x) if x2 2. (I) Evaluate the limit or state that it does not exist. If the limit does not exist, explain why. Show your work. (Note: Possible answers include + cor -0.) (II) (2 pts) Determine whether g has any horizontal asymptotes. If so, write the equation (or equations) of all horizontal asymptotes. (III) (4 pts) Determine whether g has any slant asymptotes. If so, write the equation (or equations) of all slant asymptotes. (IV) (2 pts) Determine whether g has any vertical asymptotes. If so, write the equation (or equations) of all vertical asymptotes.

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.

(5) Let g(x) = x2 +22/3. At which point(s) does the function have a local minimum? (B) 2 = 1. (C) = 0. g'KI= 2x+ g' ONE @ x=0 Fag=0 2x+ 20 (D) f has no local minimum. (E) None of these.

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2. Find the equation of the slant asymptote of the following function as 2 + +oo, if there is one: 2x -9 x2 + 5x x+5x 2x x Write your answer in the form y = mx + b. f(x) = 22 + x2 - 1

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