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15 Jun 2018

3. Find lim 1-sin (It) 21 1 - cos(1 – t) (a) 0 (c) (d) (e) The limit does not exist.

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Tod Thiel
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15 Jun 2018
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1. Find lim 2+0 sin 3.0 22 3.3 sin 4.r (0) - (d) 1 (e) The limit does not exist.

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

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