Chapter82005solutions - Chapter 8 Test 1. Evaluate x→1 No...

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Unformatted text preview: Chapter 8 Test 1. Evaluate x→1 No Calculators February 15, 2005 Name lim x→1 lim 2. Evaluate x→0 lim è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! 3 + 2x − 3 + x x 1 + cos Hπ xL 1 − x + ln x → x→0 lim è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! 3 + 2x − 3 + x x → 0 0 lim x→0 1 + cos Hπ xL 0 0 L → 1 − x + ln x lim x→1 −π sin HπxL −1 + 1 x → 0 0 L → lim x→1 −π2 = lim x→0 = 3. Evaluate 1 è!!!! 23 H3 + 2 x − 3 − xL = è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! xI 3 + 2x + 3 + xM = è!!!! 3 6 lim x→0 x è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! xI 3 + 2x + 3 + xM è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! I 3 + 2x − 3 + xMI 3 + 2x + 3 + xM è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! xI 3 + 2x + 3 + xM = lim x→0 cos HπxL −1 x2 → −1 π2 1 è!!!!!!!!!!!!!!!! ! è!!!!!!!!!!!! ! I 3 + 2x + 3 + xM x → 1+ 4. x→0 = lim+ Hx + sin xL x→ 0 x Hx + 2L y z j − lim i z j Hx + 2L Hx − 1L Hx − 1L Hx + 2L { k Evaluate x → 0+ x x → 1+ lim Hx + sin xLx → 0 0 x 1y j z lim i − j2 z x − 1{ kx + x − 2 = x → 1+ lim New 0 0 1 → lim + 5. Evaluate 1 − x2 H1 + cos xL x + sin x x→0 x→0 lim+ → L ln Hx + sin xL 1 x Hx + 2L Hx − 1L → x−x−2 = x → 1+ lim Hx + 2L Hx − 1L → x→ 0 −2 = −2 0+ → −∞ −∞ ∞ L lim + → 1 x + sin x → x→ 0 lim + − 2 x − 2 x cos x + x2 sin x 1 + cos x 0 2 → H1 + cos xL −1 x2 e0 = 1 x→0 lim Hcos xL x2 → 1 ∞ L → lim x→ 0 lim Hcos xL x2 New → − sec2 x 2 → − lim x→0 1 2 ln Hcos xL x2 so → e −1 2 0 0 → L 1 è!!!! e → → x→0 lim 6. è!!!!!!!!!!!!!!!!! ! ! ln H2 xL lim → è!!!! x →∞ ln x è!!!!!!!!!!!!!!!!! ! ! è!!!! Let f HxL = ln H2 xL and g HxL = ln x . Which of the following are true? Show your work. I. f = o HgL II. f = O HgL III. g = o HfL IV. g = O HfL ∞ ∞ L → x→∞ è!!!! e e 1 cos x H− sin xL 2x = lim x→ 0 − tan x 2x → 0 0 lim 7. 1 = lim è!!!!!!!!!!!!!!!!! ! ! x→∞ Hln 2 xL Evaluate → 0 1 ∞ → 0 so I 1 2 I and II Hln 2 xL −1 2 1 2x MH 1 2x L H2L = lim Hln 2 xL x→∞ −1 2 −∞ 2 2x dx ‡xe u x2 2x 2 0 + − + − dv e2 x 1 2 1 4 1 8 e2 x e2 x e2 x ‡ 4 ∞ → b→ −∞ lim A = x2 2 e2 x − 1 4 , x 2 e2 x + 1 4 Converges e2 x E 0 b = b 1 2 by i b2 j z − lim j − + ez j z −2 b −2 b b → −∞ 4 2e 4 k2e { 1 8. Evaluate 3 x2 − 2x dx 3 = A Hx − 2L + Bx 3 1 2 x−2 E dx = lim ‡ b→∞ 4 b → A= x Hx − 2L −2 3 3 dx → B= y z z { 2 3 ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ x Hx − 2L 3 2 3 = A x + b B x−2 −3 1 2 3 2 x A+B=0 lim Aln J NE b 0 3 = −2 A so = −3 9. ƒ b ƒ ƒ j lim ƒ ln i ƒ j ƒ 2 b→∞ ƒ ƒ k b−2 Evaluate ‡ 0 2 + ln 2 lim ‡ A b→∞ 4 + −3 2 x x−2 b→∞ = ln 2 , Converges = b→ → lim − ‡ 1 2 ‡ 0 1 2 H2 x − 1L b 0 1 2 3 dx H2 x − 1L 3 2 1 dx 1 2 b→ so we check the second integral, lim − ‡ u 1 2 H2 x − 1L 2 b−1 −1 1 +‡ 1 2 2 2 3 dx H2 x − 1L 3 2 1 dx Work on the first integral first, so −2 3 du = u = H2 x − 1L 3 2 b→ du = 2 dx E 1 2 du = dx 3 3 è!!!!!!!!!!!!!! ! 2b−1 + 2 1 2 lim− A 1 2 3 2 3 è!!!! u 2 b−1 −1 = 3 è!!!! u 3 2 = b→ lim− 1 2 = 3 2 = 3 è!!!! 3 3 2 b→ and adding the two together, we get → lim− A 1 2 2 b−1 3 2 + 3 è!!!! 3 3, 2 E 3 3 2 b→ Converges 3 3 è!!!! è!!!!!!!!!!!!!! ! lim+ A 3 − 2 b − 1 E 10. Evaluate u=1−x −1 2 x+4 ‡ è!!!!!!!!!!!!!! ! 1 − x2 0 2 1 = du = − 2 x dx 1 − b2 1 x lim ‡ è!!!!!!!!!!!! dx ! b → 1– 1 − x2 0 − 1 2 x+4 ‡ è!!!!!!!!!!!!!! dx ! 1 − x2 0 b 1 + du = x dx 1 4 lim – ‡ è!!!!!!!!!!!! dx ! b→ 1 1 − x2 0 so − 1 2 b and for the first integral, 1−b2 = A 2u2 1 E + 4 lim sin–1 b – b→1 − 0 = −1 2 b → 1– lim 2 lim ‡ b → 1– 1 è!!!!!!!!!!!! ! 1 − b2 + 1 è!!!! du u 1 + + 4J 4 lim @sin–1 xD – b 0 b→1 π 2 N = 1 + 2 π, Converges 11. Use the direct comparison test or the limit comparison test to determine if the following integral converges or diverges. lim A 1 dx ‡ è!!!! 3 x +4 8 x3E = 2 ∞ H1L lim H2L Consider ‡ x 8 x→∞ 3 è!!!! x +4 ∞ −1 3 1 3 è!!!! x +4 1 3 è!!!! x dx → lim ‡ x b→∞ 8 b −1 3 dx = 3 2 b 8 b→∞ b→∞ lim 3 2 b3 − 6 → ∞ 2 = 1, so by the Limit Comparison Test, 1 3 è!!!! x +4 dx N dx 1 3 2 è!!!! x 1 Diverges ≤ 2 3 è!!!! x → ≥ and by the Direct Comparison Test, Diverges 12. ‡ Evaluate 3 x2 + 4 x + 4 x3 + x ƒƒ ƒƒ ƒƒ 4 ln ƒ x ƒ ƒƒ ƒƒ ƒƒ ƒƒ → A+B = 3 = B=1 + dx = ‡ J 1 2 ‡ ‡ 3 x2 + 4 x + 4 x3 + x A x + Bx+C x2 + 1 and C=4 so 13. Evaluate ln Hx2 + 1L + 4 tan–1 x + C x2 − 2 x − 2 x3 − 1 dx ‡ A Hx2 + 1L + HB x + CL x = 3 x2 + 4 x + 4 4‡ 1 x dx + ‡ x x2 +1 dx + 4 ‡ if x = 0 → A = 4 1 +1 dx x2 A Hx2 + x + 1L + HBx + CL Hx − 1L = x2 − 2 x − 2 A + B = 1, so B=2 and −1 2x + 1 du = H2 x + 1L dx dx + ‡ 2 dx for the second integral, u = x2 + x + 1 ‡ x−1 x +x+1 1 = − ln Hx − 1L + ‡ du = − ln » x − 1 » + ln » x2 + x + 1 » + C u x3 − 1 = Hx − 1L Hx2 + x + 1L so x2 − 2 x − 2 x3 −1 if x=1 → − 2 = A − C, dx = ‡ dx x−1 x2 + x + 1 3 A = − 3, so A = − 1. so C=1 so A + Bx + C ...
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This note was uploaded on 08/29/2010 for the course MATH 44323 taught by Professor Anderson during the Spring '09 term at University of California, Berkeley.

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