Respostas - Calculo A - Cap 5 D - Flemming E Gonçalves

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423 5.14 – EXERCÍCIO – pg. 232 Determinar os seguintes limites com auxílio das regras de L´Hospital. 1- 2x − 4 0 x2 − 4x + 4 = lim = =0. lim 2 x→2 2x − 1 x→2 x − x − 2 3 2- lim −2 2x x2 −1 = lim = = −1 . 2 x + 4 x + 3 x → −1 2 x + 4 − 2 + 4 3- lim 2x + 6 6 x 2 + 6x = lim 2 = . 3 2 x →0 3 x + 14 x + 5 5 x + 7 x + 5x 4- 1 +1 2 +1 3 4x +1 2x + x −1 2 = lim1 = = = =∞. lim1 2 x→ 2 8 x − 4 x→ 2 4 x − 4 x + 1 1 4 4 0 − 8 −4 2 5- lim − 2 + 6 x − 3 x 2 − 11 6 − 2 x + 3x 2 − x3 . = lim = x →3 4 x 3 − 9 x 2 − 1 x 4 − 3x3 − x + 3 26 lim 1 1 1 x +1 = lim 3 = =− . 2 2 x → − 1 6 8x + 6 x + 6 x + 2 − 8 + 6 − 6 + 2 2 x + 2 x + 3x + 2 x − 1 7- lim 2x − 6 2 x2 − 6x + 7 = lim 2 = lim =0. 3 x →∞ 3 x + 7 x →∞ 6 x x + 7 x −1 8- lim − 15 x 2 15 5 5 − 5x3 = = = . lim x → −∞ − 6 x 2 6 2 2 − 2 x3 x → −1 x →0 4 2 6- 9- x →3 4 x → −1 x →∞ x → −∞ 3 35 x 4 140 x 3 7 x5 − 6 = lim = lim = +∞ . lim x → +∞ 8 x − 2 x → +∞ x → +∞ 4 x 2 − 2 x + 4 8 −1 + 2x 2 1 5 − x + x2 = lim = lim =− . 2 x → ∞ x → ∞ 2 − x − 2x 2 − 1 − 4x −4 10 - lim x →∞ lim ex ex ex = lim = lim = +∞ . x → +∞ 2 x x → +∞ 2 x2 lim 99 x 98 k x 99 = lim = ... = lim x = 0 . x x x →+∞ x → +∞ e e e 13 - lim 1 1 x lim x = = 1. e − cos x x →0 e + sen x 1 + 0 11 - 12 - x → +∞ x → +∞ x →0 x ( ) 1 14 - lim x 2 e x − 1 x → +∞ 0 . ∞? 424  1x  1     1  e − 2    x  e 1 −  = lim   x   = lim  x → +∞  1  x → +∞  − 2 x   2    x4  x    1 x ex 1 x4 = lim e . 2 . = lim = +∞ x → +∞ x 2 x x →+∞ 2 1 x 15 - limπ x→ 2 cos x ( x − π / 2) 2 = limπ x→ 2 − sen x π  2 x −  2  2 = −1 =∞. 0 2x 16 - lim x x →∞ 2 − 1 = lim x →∞ 2 x ln 2 =1. 2 x ln 2 1   1 17 - lim  −  x→2 2 x − 4 x −2   x − 2 − 2x + 4  −x+2  = lim = lim  x → 2 (2 x − 4 ) ( x − 2 )   x →2 (2 x − 4 ) ( x − 2 ) −x+2 −1 −1 = lim 2 = lim = =∞ x→2 2 x − 8 x + 8 x→2 4 x − 8 0 18 - x   lim  ln   x + 1 x → +∞ = ln lim x → +∞ 19 - x 1 = ln lim = ln 1 = 0 . x → +∞ x +1 1  x π   lim  − x →π / 2 cot g x 2 cos x   425  2 x cos x − π cot g x   = lim  x →π / 2 cot g x 2 cos x   cos x    2 x cos x − π  senx  = lim  x →π / 2  cos x  2 cos x   senx    2 xsenx − π   2 x cos x + 2 senx  = lim  = lim    = − 1. x →π / 2 − 2 senx  2 cos x  x →π / 2   20 - lim tgh x x → +∞ 1 ex − x e x − e−x e = lim x = lim x → +∞ e + e − x x → +∞ 1 ex + x e 2x e −1 x e2x − 1 = lim 2ex = lim 2 x x → +∞ e + 1 x → +∞ e + 1 ex e2x . 2 = lim 2 x = 1 x → +∞ e . 2 21 - lim x →0 lim x →0 22 - lim x→∞ senh x sen x cosh x 1 = =1 cos x 1 ln x 3 x = lim 1 x 3 3 x2 = lim x →∞ x 1 −32 x 3 2 −1 3 x3 2 = lim 3 = lim 3 = 0 x →∞ x → ∞ 1 x x →∞ 23 - sec 2 x − 2tg x x →π / 4 1 + cos 4 x lim 426 = lim x →π / 4 2 sec 2 x tg x − 2 sec 2 x − sen 4 x . 4 2 sec 2 x . sec 2 x + tg 2 x . 4 sec 2 x − 4 sec 2 x tg x 8 1 = lim = = x →π / 4 − 16 cos 4 x 16 2 24 - lim x →0 cosh x − 1 1 − cos x = lim x →0 senh x cosh x 1 = lim = =1 x → 0 + sen x cos x 1 25 - lim (1 − cos x) cot g x x →0 = lim x →0 = lim x →0 1 − cos x 1 − cos x = lim x → 0 1 tg x cot g x sen x 0 = =0 sec 2 x 1 26 - lim [ln x ln( x − 1)] x →1 1 ln( x − 1) = lim = lim x − 1 x →1 x →1 1 1 − ln x x (ln x )2 2 2 (ln x ) x 1 (ln x ) . = − lim x →1 x − 1 x →1 1 x −1 − x (ln x )2 1 + x 2 ln x 1 x = − lim (ln x )2 + 2 ln x = 0 = − lim x →1 x →1 1 = lim   1 1 27 - lim  −  x →1 2 1 − x 3 1− 3 x   ( ) ( ) 427  3 − 33 x − 2 + 2 x  = lim   3 x →1  6 1− x 1− x  1 1 − 3x1 / 3 + 2 x1 / 2 = lim 6 x →1 1 − x 1 / 3 − x1 / 2 + x 5 / 6 1 − x − 2 / 3 + x −1 / 2 = lim 1 1 5 6 x →1 − x − 2 / 3 − x −1 / 2 − x −1 / 6 3 2 6 2 −5 / 3 1 −3 / 2 x − x 1 3 2 lim 6 x →1 2 −5 / 3 1 −3 / 2 5 − 7 / 6 x + x − x 9 4 36 1 = 12 ( 28 - lim x )( ) 3 4 + ln x x →0 + 3 3 ln lim+ ln x 4+ ln x = lim+ ln x 4+ ln x x →0 x →0 = lim+ x →0 3 ln x 4 + ln x 1 3 3   3 ln x = lim+ = lim+ x = 3 ⇒  lim+ x 4+ ln x = e 3  x →0 4 + ln x x →0 1  x →0  x 29 - lim x sen x x →0 + 428 ln lim+ x sen x = lim+ ln x sen x x →0 x →0 = lim+ sen x . ln x = lim+ x →0 x →0 ln x 1 sen x 1 −1 x = lim+ = lim+ x →0 − cos sec x . cot g x x →0 x cos sec x . cot g x − sen 2 x = lim+ x →0 x cos x = lim+ x →0 − 2 sen x cos x 0 = =0 − x sen x + cos x 1 ∴ lim+ x sen x = e 0 = 1 x →0 1 30 - lim x 1− x x →1 ln lim ln x 1 1− x = lim ln x x →1 1 1− x x →1 = lim x →1 1 ln x 1− x 1 ln x = lim = lim x = −1 x →1 1 − x x →1 − 1 1 1 ∴ lim x 1− x = e −1 = x →1 e 31 - lim− (1 − x) cos πx 2 x →1 ln lim− (1 − x) cos x →1 = lim− cos x →1 πx 2 πx 2 = lim− ln(1 − x) cos πx 2 x →1 . ln (1 − x) = lim− x →1 ln (1 − x) 1 πx cos 2 −1 −1 1− x = lim− = lim x →1 π x π x x→1− sec tg (1 − x )sec π x tg π x 2 2 2 2 429 −1 = lim− x →1 sen 1 πx 2 πx πx cos cos 2 2 πx πx πx π sen cos 2 − 2 cos 2 2 2 2 = lim− = lim =0 x →1 π x x→1− π πx πx (1 − x ) sen (1 − x ) cos − sen 2 2 2 2 (1 − x ) Assim, lim− (1 − x) . cos πx 2 = e 0 = 1. x →1 32 - lim x sen π / x x → +∞ π  −π   cos  2 x x x = lim  x → +∞ 1 −1 x x2 sen = lim x → +∞ π = lim π cos x → +∞ π =π x x2/3 33 - lim 2 x → ∞ ( x + 2)1 / 3 1 1 3 1  x2   2x  3  =  lim =  lim 2 = 13 = 1   x → ∞ x + 2   x →∞ 2 x  34 - lim x →∞ senh x x = lim x→∞ cosh x =∞ 1 35 - lim (2 x − 1) 2 / x x→∞ ln lim (2 x − 1) 2 / x = lim ln (2 x − 1) 2 / x x →∞ x →∞ 2 ln (2 x − 1) 2 ln (2 x − 1) = lim x →∞ x x →∞ x 2 2 4 = lim 2 x − 1 = lim =0 x →∞ x → ∞ 1 2x − 1 = lim 2 ⇒ lim (2 x − 1) x = e0 = 1 x →∞ 430 3 2 36 - lim (cos 2 x) x x→0 3 2 3 2 ln lim (cos 2 x) x = lim ln(cos 2 x) x x →0 x →0 − sen 2 x . 2 cos 2 x 3 = lim 2 ln (cos 2 x) = lim x →0 x x →0 2x − 6 sen 2 x 1 − 6 . 2 cos 2 x = lim = lim x →0 cos 2 x 2 x x →0 2 (cos 2 x ) + 2 x (− sen 2 x ) 2 3 = lim x →0 − 12 cos 2 x − 12 = = −6 2 cos 2 x − 4 x sen 2 x 2 − 0 3 ⇒ lim (cos 2 x ) x 2 = e − 6 = x →0 37 - lim+ x →0 1 e6 ln(sen a x) ln(sen x) a cos ax sen ax a cos ax sen x = lim+ = lim+ x →0 x →0 cos x sen ax cos x sen x = lim+ a cos ax cos x − a 2 sen x . sen ax − sen ax . sen x + a cos x . cos ax = lim+ a−0 =1 0+a x →0 x →0 5  1  38 - lim  − 2  x →3  x −3 x − x −6  x+2−5   1  5 x−3 1  = lim  = lim  = lim  − = x →3  x − 3 ( x − 3)( x + 2)  x→3  ( x − 3) ( x + 2 )  x →3 ( x − 3) ( x + 2 ) 5 39 - lim x→0 + 1 x tgx ln lim+ x −tgx = lim+ ln x −tgx = lim+ − tgx ln x = lim+ − x →0 = lim+ − x →0 x →0 x →0 ln x 1 = lim+ = lim cot g x x →0 x cos sec 2 x x →0 + x →0 1 x sen 2 x ln x 1 tg x 431 2 sen x cos x 0 sen 2 x = lim+ = =0 0 x →0 x → x 1 1 1 ⇒ lim+ tgx = e 0 = 1 x →0 x = lim+ 40 - lim x 2 2 + ln x x →0 + ln lim+ x 2 2 + ln x = lim+ ln x x →0 2 2 + ln x x →0 1 2 2 ln x 2 = lim+ ln x = lim+ = lim+ x = 2 x →0 2 + ln x x →0 2 + ln x x →0 1 x 2 ⇒ lim+ x 2+ ln x = e 2 x →0 41 - limπ (1 − tg x ) sec 2 x x→ 4 = limπ x→ 4 1 − tg x 1 − tg x = limπ x → 4 cos 2 x 1 sec 2 x 2  2  −  2 sec x 4 1 2  = . =1 = limπ = x → 4 − 2 sen 2 x −2 2 2 42 - lim x→∞ x ln x x + ln x 1 + ln x 1 + ln x x = lim =∞ x→∞ 1 1 1+ 1+ x x x = lim x →∞ 43 - lim (e x + x)1 / x x →0 ln lim (e x + x)1 / x = lim ln(e x + x)1 / x = lim x →0 x →0 x e +1 x ex +1 = lim e + x = lim x x →0 x →0 e + x 1 1+1 2 = = =2 1+ 0 1 1 ⇒ lim (e x + x) x = e 2 x →0 x→0 ln (e x + x) 1 ln (e x + x) = lim x →0 x x