MATH 41 Midterm: MATH 41 Stanford 11Exam1sol

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turquoisegnu128

15 Feb 2019

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Solutions to math 41 exam 1 october 20, 2011: (16 points) find each of the following limits, with justi cation. If the limit does not exist, explain why. If there is an in nite limit, then explain whether it is or . (a) lim x ( x + x)2. 1 + x x lim x x + 2x x + x2. = lim x (cid:18) x + 2x x + x2. 1 x + 2 x2 + 1 x1/2. Hence, the limit is of the form 1 both the numerator and denominator a positive then, 0 , which implies that the limit is either or . 1 + x x x + 2x x + x2 x x + 2x x + x2. = lim x because the numerator goes to and the denominator goes to 1. cos(x) + sin(x) + 1 (b) lim x 3 .

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3. (21 pts) Evaluate the limit or say that the limit does not exist. If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L'Hospitals's Rule to justify your answer. (Note: Possible answers include + or -..) (a) (7 pts) x2 - 2x - 3 - lim x+3 Vx+1 - 2 x2 - 2x - 3 Vx+ 1 + 2 lim x+3 Vx+1 -2 Vx+1 +2 (x - 3) (x + 1) (Vx+ = lim x+3 x + 1 - 4 = lim x+3 (x - 3) (x + 1) (Vx+1 +2). -- x-3 = lin (x + 1) (Vx+1 +2) = (3 + 2) (V3+1 +2) = = 4 (V4 +2) = 16 (b) (7 pts) lim xsin (In x) X >0+ NOTE: We cannot apply the Product Law, since lim sin (1n x) DOES NOT EXIST. On the other hand, -1s sin (in x) s 1 and , therefore, -xs x sin (in x) sx, (since x > 0). Since lim X = lim (-X) = 0, x0+ X0 the inequality above implies that lim xsin (in x) = 0, by the Squeeze Theorem . x0+ EXPLANATION : Since lim ln x = -0, x →0+ sin (lnx) oscillates between 1 and -1 as x + 0*, and, therefore lim sin (In x) DOES NOT EXIST. x >0+ (c) (7 pts) in x+2- (lnx) +In 20 -= -00 (x - 2) 70-

4. (26 pts) Evaluate the limit or say that the limit does not exist . If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L' Hospitals's Rule to justify your answer. (Note: Possible answers include + or -.) X-2 (a) (10 pts) lim – x+2 782 +5 - 3 (b) (8 pts) xvot COS X OSCILLATES lim x → x + 4 x2 Since -1 scos x si, we have that -1 (c) (8 pts) cos x x + 4 x2 1 x + 4 x? x + 4 x2 -1 1 and since lim – Elim- x+ x + 4 x2 + x + 4 x2 * +422=0, it follows by the Squeeze Theorem that lim Co2 = 0. X +0 x + 4 x 2

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.

MATH 41 Midterm: MATH 41 Stanford 11Exam1sol

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