Respostas - Calculo A - Flemming E Gonçalves - Cap 3 C

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164 3.10 – EXERCÍCIO – pg. 83 1 – Para cada uma das seguintes funções, ache lim x→2 f ( x ) − f ( 2) : x−2 (a) f ( x) = 3 x 2 3 x 2 − 12 ( x − 2)(3 x + 6) = lim = 12 x→2 x→ 2 x−2 x−2 lim 1 ,x≠0 x 1 1 2−x − − ( x − 2) 1 −1 lim x 2 = lim 2 x = lim ⋅ = x→2 x − 2 x→2 x − 2 x→2 2x x−2 4 (b) f ( x) = 2 2 x 3 4 2 2 2 8 ( x − 2) x +  x − 3 4 4 8 3 3 = lim = + = lim 3 x→2 x→2 x−2 x−2 3 3 3 (c) f ( x) = (d) f ( x) = 3 x 2 + 5 x − 1 3 x 2 + 5 x − 1 − 21 3 x 2 + 5 x − 22 ( x − 2)(3 x + 11) = lim = lim = 17 x→2 x→2 x→2 x−2 x−2 x−2 lim 1 , x ≠ −1 x +1 3 − x −1 1 1 − 3( x + 1) − 2 + x −1 −1 lim x + 1 3 = lim = lim ⋅ = x→2 x→2 x → 2 3( x + 1) x − 2 x−2 x−2 9 (e) f ( x) = (f) f ( x) = x 3 x3 − 8 ( x − 2)( x 2 + 2 x + 4) = lim = 12 x→2 x − 2 x→2 x−2 lim 2 -
Esboçar o gráfico das seguintes funções e dar uma estimativa dos limites indicados 165 (a) f ( x) = x2 − 9 ; x−3 lim f ( x) = 6 x →3 y 8 7 6 5 4 3 2 1 x -4 -3 -2 -1 1 2 3 4 2 3 4 -1 -2 -3 (b) f ( x) = x3 − 3x + 2 ; x2 − 4 lim f ( x) = − x → −2 9 . 4 y 8 7 6 5 4 3 2 1 x -4 -3 -2 -1 1 -1 -2 -3 (c) f ( x) = 3 x −1 ; x −1 lim f ( x) = x →1 3 . 2 y 2 1 x 1 (d) f ( x) = x −1 ; x3 − 1 2 1 lim f ( x) = . x →1 3 3 4 166 y 2 1 x -2 -1 1 2 3 – Calcular os limites indicados no Exercício 2 e comparar seus resultados com as estimativas obtidas. x2 − 9 ( x − 3)( x + 3) = lim = lim( x + 3) = 6. x →3 x − 3 x →3 x →3 x−3 (a) lim x3 − 3x + 2 ( x + 2)( x − 1) 2 ( x − 1) 2 9 = lim = lim =− . (b) lim 2 x → −2 x → − x → − 2 2 x −4 x−2 ( x + 2)( x − 2) 4 (c) lim 3 x →1 x −1 u3 − 1 (u − 1)(u 2 + u + 1) (u 2 + u + 1) 3 = lim 2 = lim = lim = . u →1 (u + 1) 2 x − 1 u →1 u − 1 u →1 (u − 1)(u + 1) x −1 x −1 1 = lim = . 3 2 x →1 x − 1 x →1 ( x − 1)( x + x + 1) 3 (d) lim Nos exercícios de 4 a 27 calcule os limites. x3 +1 ( x + 1)( x 2 − x + 1) (−1) 2 − (−1) + 1 3 lim = = =− . 2 x → −1 x − 1 x → −1 ( x − 1)( x + 1) 2 −1−1 4 - lim t 3 + 4t 2 + 4t (t + 2)(t 2 + 2t ) 4−4 0 = lim = = = 0. 5 - lim t → −2 (t + 2)(t − 3) t → −2 (t + 2)(t − 3) −2−3 −5 x 2 + 3x − 10 x+5 ( x − 2)( x + 5) 2+5 7 = lim = lim = = = 1. 6 - lim 2 x→2 3x − 5 x − 2 x → 2 ( x − 2)(3 x + 1) x→2 3x + 1 3⋅ 2 +1 7 7 - lim t→ 5 2 2t 2 − 3t − 5 (2t − 5)(t + 1) 5 7 = lim = +1 = . 5 2t − 5 (2t − 5) 2 2 t→ 2 167 8 - lim x→a x 2 + (1 − a ) x − a ( x − a )( x + 1) = lim = a +1. x → a x−a ( x − a) 3 x 2 − 17 x + 20 ( x − 4)(3 x − 5) 12 − 5 7 9 - lim 2 = lim = = = 1. x → 4 4 x − 25 x + 36 x → 4 ( x − 4)( 4 x − 9) 16 − 9 7 x2 + 6 x + 5 ( x + 1)( x + 5) − 1 + 5 4 = lim = =− . 2 x → −1 x − 3 x − 4 x → −1 ( x + 1)( x − 4) −1− 4 5 10 - lim x2 −1 ( x − 1)( x + 1) − 1 − 1 − 2 = lim = = = −2 . 2 x → −1 x + 3 x + 2 x → −1 ( x + 1)( x + 2) −1+ 2 1 11 - lim 12 - lim x2 − 4 ( x − 2)( x + 2) = lim = 4. x − 2 x→2 ( x − 2) 13 - lim x 2 − 5x + 6 ( x − 2)( x − 3) 2 − 3 −1 1 = lim = = = . 2 x → 2 ( x − 2)( x − 10) 2 − 10 − 8 8 x − 12 x + 20 x→2 x→2 (2 + h) 4 − 16 h 4 + 8h3 + 24h 2 + 32h + 16 − 16 h/ (h3 + 8h 2 + 24h + 32) lim = lim = lim 14 - h → 0 h →0 h→0 h h h/ = 32 15 - lim t →0 (4 + t ) 2 − 16 16 + 8t + t 2 − 16 t/ (8 + t ) = lim = lim = 8. t →0 t →0 t t t/ 16 - lim 25 + 3t − 5 25 + 3t/ − 25 3 = lim = . t → 0 t ( 25 + 3t + 5) t 10 / 17 - lim a 2 + bt − a a 2 + bt/ − a 2 b , a>0. = lim = t →0 t t/ ( a 2 + bt + a ) 2a 18 - lim h −1 h −1 1 = lim = . h → 1 h −1 (h − 1)( h + 1) 2 t →0 t →0 h →1 19 - lim h → −4 2( h 2 − 8 ) + h 2(h 2 − 8) − h 2 2h 2 − 16 − h 2 = lim = lim = h → −4 h+4 (h + 4)( 2(h 2 − 8) − h) h→−4 (h + 4)( 2(h 2 − 8) − h) = lim h → −4 (h + 4)(h − 4) (h + 4)( 2(h 2 − 8 − h) = −8 = −1 . 8 168 3 20 - lim h →0 8+h −2 h u3 = 8 + h ⇒ h = u3 − 8 u−2 u−2 1 lim 3 = lim = . 2 u→2 u − 8 u → 2 (u − 2)(u + 2u + 4) 12 −1 1+ x −1 1/ + x/ − 1/ = lim = . x → 0 −x 2 − x/ ( 1 + x + 1) 21 - lim x →0 x2 + a2 − a 22 - lim 2 x →0 2 = lim x +b −b 3 23 - lim x→a x →0 ( x 2 + a 2 − a 2 )( x 2 + b 2 + b) 2 2 2 2 2 ( x + b − b )( x + a + a ) 2/ b b = , a, b > 0 . x →0 2 /a a = lim x −3 a x−a Fazendo: x = u3 a = b3 com b ≠ 0 e a ≠ 0 temos: u −b u −b 1 1 = lim = 2 = 2. 3 3 2 2 2 2 u →b u − b u → b (u − b)(u + bu + b ) b +b +b 3b 1 = . 33 a 2 lim 3 24 - lim x −1 x →1 4 x −1 Fazendo x = u 12 3 lim x →1 4 3 25 - lim x →1 x −1 x −1 3 x →1 u 4 −1 (u − 1)(u 3 + u 2 + u + 1) 4 = lim = . u →1 u 3 − 1 u →1 3 (u − 1)(u 2 + u + 1) = lim x 2 − 23 x + 1 ( x − 1) 2 Fazendo lim , u ≥ 0 temos: 3 x = u , temos: x 2 − 23 x + 1 u 2 − 2u + 1 (u − 1)(u − 1) 1 = lim = lim = . 2 3 2 2 2 2 u →1 (u − 1) u →1 (u − 1) (u + u + 1) 9 ( x − 1) 169 − 2 −1 3− 5+ x (9 − 5 − x)(1 + 5 − x (4 − x)(1 + 5 − x ) . = lim = lim = = x→4 1 − 5 − x x → 4 (1 − 5 + x )(3 + 5 + x x → 4 ( −4 + x )(3 + 5 + x ) 6 3 26 - lim 27 - lim x →0 1+ x − 1− x 1/ + x − 1/ + x 2x 2 = lim = lim = = 1. x → 0 x → 0 x x ( 1 + x) + 1 − x ) x( 1 + x) + 1 − x ) 2