MATH 41 Midterm: Stanford MATH 41 exam1sol

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turquoisegnu128

7 Mar 2019

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Solutions to math 41 first exam october 15, 2013: (16 points) find each of the following limits, with justi cation. If the limit does not exist, explain x2 ln(x/2) x2 1 why. If there is an in nite limit, then explain whether it is or . (a) lim x 1+ (4 points) by substituting x = 1 we get ln(1/2) in the numerator, which is nonzero, and 0 in the denominator. So it is enough to nd its sign. Observe that when x 1+ we get x2 ln(x/2) ln(1/2) < 0 and x2 1 > 0. So lim x 1+ x2 ln(x/2) x2 1 z2 4 z2 + z 2 (b) lim z 2 (4 points) direct substitution gives the indeterminate form 0 nipulation: So we need some algebraic ma- lim z 2 z2 4 z2 + z 2. = lim z 2 (z + 2)(z 2) (z + 2)(z 1) z 2 z 1.

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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.