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2 Jul 2018

1 tan (2) 2. (a) lim 2+2 x2 +1 5 sin(3.c) – 3 (b) lim 20 T (e) lim (3z* sin (-)] = 0

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Casey Durgan
Casey DurganLv2
2 Jul 2018
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5. Which of the below limits are equal to on? etetve 2+2+ x2 – 2.r ) X X -2 tve te B. lim tan - 2 + /2+ V tye 22 - 25 C +5 lim lin ( 2+3+ 22 - 3 *337 Xlx-3) Dling – In(sin x) lim 22 - 25 r2- 3. - lih +3 lim ks)

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-

11. Let f(1) = cos 1. Use the definition of the derivative to prove that f'(:1) = -sin . Solution: We have · f'(:1) = lim cos(2+h) - cos(2) +0 cos(1) cos(h) - sin() sin(h) - cos(1) = lim h0 sin / – Jin contar) (cos(1) = -1) = sin(e) (sinm) h0 sin(h) cos(h) - = lim cos(2) h cos(h) - 1 = cos(x) lim h- oh = cos(x)0 - sin(x)(1) = - sin h c) lim 0 / h

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