MATH 41 Midterm: MATH 41 Stanford Exam1Solutions

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turquoisegnu128

15 Feb 2019

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Solutions to math 41 first exam october 14th, 2014: (25 points) compute the following limits. If the limit does not exist, explain why. Since sin2(x) 0, we get the inequalities. 2 sin2(x) sin2(x) cos(1/x) 2 sin2(x), x 6= 0 since limx 0. 2 sin2(x) = 0, by the squeeze theorem we nd (sin2 x)2cos(1/x) = 0. lim x 0 (b) lim t 0(cid:18) We do the following algebraic manipulation: t 0(cid:18) lim. 1 1 + t t 1 + t. 1 t (1 1 + t)(1 + 1 + t) (t 1 + t)(1 + 1 + t) Expanding the numerator we nd the limit equals. 1 (1 + t) t 1 + t(1 + 1 + t) T t 1 + t(1 + 1 + t) The nal expression is continuous at t = 0, so we may evaluate, nding t 0(cid:18) lim.

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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

Hello, I am having an extremely diffucult time understanding trig subs in Calc2. Can someone please fully expalin to me why the limits of integration change from 3sqrt3/2 to pi/3?....and than change once again from 1 to 1/2. I know you can plug the number somewhere to see if they change, but I'm not sure where to do this...or even WHY to do this? I still don't understand why we change limits of integration in other calc problems like u-sub. Can you please explain it like you're explaining it to someone who has no math skills. I really appreiciate a clear explanation without too much math terminology.

God bless and thank you!

MATH 41 Midterm: MATH 41 Stanford Exam1Solutions

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