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13 Dec 2017

7. If lim f(x) = 3, lim g(x) = -2, and lim h(x) = 4, evaluate the limit (C) 13 (A) -1 (D) 5 (B) 3 (E) 7

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Irving Heathcote
Irving HeathcoteLv2
16 Dec 2017
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[ -2 e* if x< 0 3. (24 pts) Let g(x) = 1 X+6 if x20. [x - 3 (I) Evaluate the limit. (Note: Possible answers include + or - .) [You may not use a table of values, a graph, or L'Hospitals's Rule to justify your answer. ] SHOW YOUR WORK. (a) (3 pts) lim g(x) = ; (b) (3 pts) lim g(x) = x 10 x-10+ (c) (3 pts) lim g(x) = ; (d) (3 pts) lim g(x) = 14 = * 3+ (e) (3 pts) lim g(x) = ; (f) (3 pts) lim g(x) = X + (II) (6 pts) Using the definition of continuity, show that gis continuous at 0. SHOW YOUR WORK.

5. If f(x)= 27° -5x + 3x –7 then A) lim f(n= f(x)=0 B) lim f(x)=0 D) lim f(x)=-- C) lim E) None of these

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-

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