MATH 41 Final: MATH 41 Stanford 12Finalsol

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turquoisegnu128

15 Feb 2019

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Solutions to math 41 final exam december 10, 2012: (10 points) find each of the following limits, with justi cation. If there is an in nite limit, then explain whether it is or . 0 ln(t + 1) dt (a) lim x2 x 0r x continuous and 02 = 0 and r 0. 0 ln(t + 1)dt (5 points) we are dealing with a 0 lim x 0r x x2. 0 indeterminate, since the numerator and denominator are. We can apply l"hospital: (now this is again indeterminate 0. 0 ) (using l"hospital again) ln(x + 1) Note that in the rst equality we used ftc: d. 0 ln(t + 1)dt = ln(x + 1). dxr x (b) 1/x (1 + sin x) lim x 0+ (5 points) note that this is an indeterminate limit of form 1 . So, let l = lim x 0+ will compute ln l: (1 + sin x)

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.