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Chapter82004solutions

# Chapter82004solutions - Chapter 8 Test 1 lim Evaluate cos t...

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Unformatted text preview: Chapter 8 Test 1. lim Evaluate cos t − 1 et − t − 1 Evaluate x−4 3 è!!!!!!!!!!!! ! x+4 −2 2 No Calculators February 24, 2004 Name t→0 lim cos t − 1 et − t − 1 so L→ t→0 t→0 → lim 0 0 lim − sin t et − 1 → 0 0 so L→ t→0 lim − cos t et = −1 1 = −1 2. x→4 x→4 lim → 0 0 x−4 3 è!!!!!!!!!!!! ! x+4 −2 so L → lim = 3. 2 3 è!!!!!!!!!!!! ! Hx − 4L IHx + 4L 3 + 2 x + 4 + 4M = lim 2 3 3 è!!!!!!!!!!!! ! è!!!!!!!!!!!! ! x→4 I x + 4 − 2M IHx + 4L 3 + 2 x + 4 + 4M x→4 1 3 x → 0+ NEW L→ 4. lim Htan xLx → 00 x→0 Hx + 4L 3 + 2 Evaluate 3 è!!!!!!!!!!!! ! x+4 +4 Hx + 4L = 12 1 −2 3 = = x → 0+ lim+ x→0 lim+ ln Htan xL 1 x lim Htan xLx → → 1 4+4+4 Hx − 4L IHx + 4L 3 + 2 2 = 3 H4L = 12 OR 3 è!!!!!!!!!!!! ! x + 4 + 4M Hx + 4L − 8 −∞ ∞ 0 1 so → L→ e0 x→0 lim+ 1 1 tan x sec2 x −1 x2 = x→0 lim+ − x2 cos x sin x cos x → 0 0 −2 x cos2 x − sin x x → 1− 2 = Evaluate 1 x → 1− NEW 5. lim xI 1 − x M x → 1− → ln x 1∞ lim xI 1 − x M 0 0 1 x lim 1−x → so L→ Evaluate x→1 x→1 1 1y z j − lim i z j x − 1{ k ln x = lim x→1 1 1y z j − lim i z j x − 1{ k ln x → ∞−∞ 0 0 1 2 → = so x → 1− lim −1 = −1 → e−1 = 1 e L → lim 6. x→1 x→∞ lim so 7. f = o HgL Let f HxL = ln Hln xL and g HxL = ln x3 . 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 ln Hln xL ln x3 = x→∞ Hln xL Hx − 1L 1 1 + ln x + 1 x − 1 − ln x L → lim x→1 1 x Hx − 1L + ln x 1− 1 x = x→1 lim Hx − 1L + x ln x x−1 → 0 0 lim and π 2 Evaluate cos x ‡ è!!!!!!!!!!!!!!!! !!!!!! dx 1 − sin x 0 f = O HgL ln Hln xL 3 ln x → ∞ ∞ so L→ x→∞ lim 1 ln x 3 x 1 x → O so I and II are true = t→H t→H lim π 2 L L − = lim π 2 cos x ‡ è!!!!!!!!!!!!!!!! !!!!!! dx 1 − sin x 0 1 − sin t π 2 u = 1 − sin x Au 2 1 and − du = cos x dx − 8. Evaluate ‡ 1 u −1 2 du = −2 9. 1 1y z dx j = lim ‡ i − z j t→∞ x + 3{ kx + 2 t 0 1y i1 z dx − z j ‡j x + 3{ kx + 2 ∞ 0 t→H lim π 2 1 − sin t 1 L − E = 2 = Evaluate −∞ t → −∞ lim ‡ t 1 = 3 2 t → −∞ lim 1 i 1 dxy z j z j 2 H x L2 + 1 k 2 { 2 3 ‡ t 2 1 2 ‡ 1 3 x2 + 4 dx ƒx+2ƒt ƒ ƒ ƒ ƒ lim Aln ƒ ƒ ƒ ƒx+3ƒE ƒ ƒ ƒ0 t→∞ ƒ ƒ = ƒt+2 ƒ ƒ lim ln ƒ ƒ ƒt+3 ƒ t→∞ ƒ ƒ ƒ ƒ − ln 2 ƒ ƒ ƒ ƒ 3 ƒ = ln 3 2 u= x 2 and du = 1 2 dx 1 u2 +1 ‡ 1 du = 3 2 t → −∞ 2 10. Evaluate 3 ‡ è!!!!!!!!!!!!!! dx ! x2 − 1 1x = 11. 3J π 3 2 = c → 1+ Use the direct comparison test or the limit comparison test to determine if the following integral converges or diverges. 1 dx ‡ è!!!! x + cos x 0 1 ‡ è!!!! dx x 0 π 2 π 2 − 3 sec−1 1N lim ‡ c 3 è!!!!!!!!!!!!!! dx ! x x2 − 1 2 lim @tan−1 u D 1 2 t 2 = 3 2 tan−1 1 2 − 3 2 t → −∞ lim tan−1 t 2 = 3 2 tan−1 1 2 + 3π 4 = π 3 è!!!!!!!!!!!!!! dx ! x x2 − 1 = c → 1+ lim @3 sec−1 x D 2 c = 3 sec−1 2 − c → 1+ lim sec−1 c Direct C → 12. ‡ Evaluate 2x + 1 è!!!! x + cos x 1 ‡ x2 ≤ 2x + 1 A − 7 x + 12 1 è!!!! x and = dx B 2 lim+ Ax 2 E 1 π 2 t→0 = t è!!!!!!! ! 2π so it converges − 2x + 1 x=4 x=3 x2 13. Evaluate N dx x−3 x−4 7 x + 12 = A Hx − 4L + B Hx − 3L → 9=B → −7 = A → − 7 ln » x − 3 » + 9 ln » x − 4 » + C dx = + ‡ 3 x2 − 2 x + 12 Hx2 + 4L2 dx ‡J Cx + D i Ax + B j + j ‡j 2 Hx2 + 4L Hx2 + 4L2 kx + 4 3 x2 − 2 x + 12 = HAx + BL Hx2 + 4L + HCx + DL 3=B 0=A −2 = 4 A + C → C = −2 12 = 4 B + D → D=0 3 2x x 3 dx = tan−1 dx − ‡ =‡ 2+ 4 2 x 2 2 Hx2 + 4L ‡ 3 x2 − 2 x + 12 2 dx = y z dx z z { + Hx2 + 4L −1 +C ...
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