Prévia do material em texto
LISTA DE EXERCÍCIOS DE CÁLCULO I (LIMITES) Observações: • Os limites estão desenvolvidos na parte final da lista; • Não tente decorar as soluções pois você estará estudando errado! Tende entende-‐las e, em caso de dúvidas, consulte o monitor ou professor. • Procure agrupar os procedimentos de solução por similaridade. 1). lim x→2 x2 − 7x +10 x2 − 4 2). lim x→−1 x2 + x − 2 x2 −1 3). lim x→5 x2 + 2x − 35 x2 −10x + 25 4). lim x→25 5− x 25− x 5). lim x→9 9 − x x − 3 6). lim x→0 (x + 3)3 − 27 x 7). lim x→0 x2 x2 +12 − 12 8). lim x→0 3 x 1 5+ x − 1 5− x ⎛ ⎝⎜ ⎞ ⎠⎟ 9). lim x→4 (x − 4)3 4 − x 10). lim x→0 xsen(x) x 11). lim x→3 5x2 −8x −13 x2 − 5 l2). lim x→2 3x2 − x −10 x2 − 4 13). lim x→3 x4 −81 2x2 − 5x − 3 14). lim x→−2 1 x + 1 2 x3 + 8 15). lim x→4 3− x + 5 x − 4 16). lim x→27 x − 27 x 1 3 − 3 17). lim x→0 sen(5x) 3x 18). lim x→0 x3 − 7x x3 19). lim x→0 x4 + 5x − 3 2 − x2 + 4 20). lim x→1 x3 −1 (x −1)2 21). lim x→∞ 100 x2 + 5 22). lim x→−∞ 7 x3 − 20 23). lim x→∞ 3x3 −1000x2 24). lim x→−∞ x4 + 5x2 +1 25). lim x→∞ x5 − x2 + x −10 26). lim x→−∞ x + 7 3x + 5 27). lim x→∞ 7x2 + x −100 2x2 − 5x 28). lim x→∞ x2 − 3x + 7 x3 +10x − 4 29). lim x→−∞ 7x2 + x +11 4 − x 30). lim x→∞ x3 + 7x 4x3 + 5 31). lim x→∞ x − x2 + 7 32). lim x→−∞ x − x2 + 7 33). lim x→∞ x + 3 9x2 + 5x 34). lim x→−∞ x + 3 9x2 + 5x 35). lim x→∞ log x 6 − 500 x6 + 500 ⎛ ⎝⎜ ⎞ ⎠⎟ SOLUÇÕES: 1). lim x→2 x2 − 7x +10 x2 − 4 = limx→2 (x − 2)(x − 5) (x − 2)(x + 2) = limx→2 (x − 5) (x + 2) = − 3 4 2). lim x→−1 x2 + x − 2 x2 −1 = limx→−1 (x + 2)(x −1) (x −1)(x +1) = nãoexiste 3). lim x→5 x2 + 2x − 35 x2 −10x + 25 = limx→5 (x + 7)(x − 5) (x − 5)2 = limx→5 x + 7 x − 5 = nãoexiste 4). lim x→25 5− x 25− x = limx→25 5− x 25− x × 5+ x 5+ x = lim x→25 25− x (25− x)(5+ x ) = limx→25 1 5+ x = 1 10 5). lim x→9 9 − x x − 3 = limx→9 9 − x x − 3 × x + 3 x + 3 = lim x→9 (9 − x)( x + 3) x − 9 = −6 6). limx→0 (x + 3)3 − 27 x = limx→0 (x + 3)2(x + 3)− 27 x = lim x→0 x3 + 9x2 + 27x + 27− 27 x = limx→0 x3 + 9x2 + 27x x = lim x→0 x(x2 + 9x + 27) x = limx→0 x 2 + 9x + 27 = 27 7). lim x→0 x2 x2 +12 − 12 = lim x→0 x2 x2 +12 − 12 × x 2 +12 + 12 x2 +12 + 12 = lim x→0 x2 ( x2 +12 + 12) (x2 +12)−12 = lim x→0 x2 ( x2 +12 + 12) x2 = lim x→0 x2 +12 + 12 = 2 12 8). lim x→0 3 x 1 5+ x − 1 5− x ⎛ ⎝⎜ ⎞ ⎠⎟ = limx→0 3 x (5− x)− (5+ x) (5+ x)(5− x) ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→0 3 x −2x 25− x2 ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→0 −6 25− x2 ⎛ ⎝⎜ ⎞ ⎠⎟ = − 6 25 9). lim x→4 (x − 4)3 4 − x = limx→4 (x − 4) 4 − x 2 4 − x = lim x→4 (x − 4) 4 − x = 0 10). lim x→0 xsen(x) x como: −1≤ xx ≤1 então: − sen(x) ≤ xsen(x)x ≤ sen(x) pelo teorema do sanduiche: lim x→0 xsen(x) x = 0 11). lim x→3 5x2 −8x −13 x2 − 5 = 8 4 = 2 l2). lim x→2 3x2 − x −10 x2 − 4 = limx→2 (x − 2)(3x + 5) (x − 2)(x + 2) = limx→2 (3x + 5) (x + 2) = 3(2)+ 5 (2)+ 2 = 11 4 13). lim x→3 x4 −81 2x2 − 5x − 3 = limx→3 (x2 − 9)(x2 + 9) (x − 3)(2x +1) = lim x→3 (x − 3)(x + 3)(x2 + 9) (x − 3)(2x +1) = lim x→3 (x + 3)(x2 + 9) (2x +1) = ((3)+ 3)((3)2 + 9) 2(3)+1 = 108 7 14). lim x→−2 1 x + 1 2 x3 + 8 = limx→−2 x + 2 2x x3 + 8 = lim x→−2 x + 2 2x 1 x3 + 8 como A3 + B3 = (A + B)(A2 − AB + B2 ) = lim x→−2 x + 2 2x(x + 2)(x2 − 2x + 4) = lim x→−2 1 2x(x2 − 2x + 4) = − 1 48 15). lim x→4 3− x + 5 x − 4 = limx→4 3− x + 5 x − 4 × 3+ x + 5 3+ x + 5 = lim x→4 9 − (x + 5) (x − 4)(3+ x + 5) = lim x→4 4 − x (x − 4)(3+ x + 5) = lim x→4 −(x − 4) (x − 4)(3+ x + 5) = −1 3+ 4 + 5 = − 1 6 16). lim x→27 x − 27 x 1 3 − 3 = lim x→27 x1 3( )3 − 33 x1 3 − 3 como A3 + B3 = (A + B)(A2 − AB + B2 ) = lim x→27 (x1 3 − 3)((x1 3)2 + 3x1 3 + 9) x1 3 − 3 = lim x→27 ((x1 3)2 + 3x1 3 + 9) = ((27)1 3)2 + 3(27)1 3 + 9 = 27 17). lim x→0 sen(5x) 3x = limx→0 5 5 sen(5x) 3x = limx→0 5 3 sen(5x) 5x como lim x→0 sen(x) x =1, então: limx→0 sen(5x) 5x =1 lim x→0 5 3 sen(5x) 5x = 5 3 (1) 18). lim x→0 x3 − 7x x3 = limx→0 x(x2 − 7) x(x2 ) = limx→0 x2 − 7 x2 = −∞ 19). lim x→0 x4 + 5x − 3 2 − x2 + 4 = −30− = +∞ 20). lim x→1 x3 −1 (x −1)2 como A3 + B3 = (A + B)(A2 − AB + B2 ) = lim x→1 (x −1)(x2 + x +1) (x −1)(x −1) = limx→1 x2 + x +1 x −1 lim x→1− (1− )2 + (1− )+1 (1− )−1 = >1 0− = −∞ lim x→1+ (1+ )2 + (1+ )+1 (1+ )−1 = >1 0+ = +∞ ⎫ ⎬ ⎪ ⎪ ⎭ ⎪ ⎪ o limite não existe! 21). lim x→∞ 100 x2 + 5 = 100 ∞ = 0 22). lim x→−∞ 7 x3 − 20 = 7 −∞ = 0 23). lim x→∞ 3x3 −1000x2 = ∞−∞ = indeterminação para remover a indeterminação, basta fazer: lim x→∞ 3x3 −1000x2 = lim x→∞ x2 (3x −1000) = (∞)(∞) = ∞ 24). lim x→−∞ x4 + 5x2 +1= (x4 )+ (5x2 +1){ } = ∞ +∞ = ∞ 25). lim x→∞ x5 − x2 + x −10 = ∞−∞ = indeterminação lim x→∞ x5 − x2 + x −10 = lim x→∞ x2 (x3 −1)+ (x −10){ } = (∞)(∞)+∞ = ∞ +∞ = ∞ 26). lim x→−∞ x + 7 3x + 5 = limx→−∞ x 1+ 7x ⎛ ⎝⎜ ⎞ ⎠⎟ x 3+ 5x ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→−∞ 1+ 7x 3+ 5x = 1+ 03+ 0 = 1 3 27). lim x→∞ 7x2 + x −100 2x2 − 5x = limx→∞ x2 7+ 1x − 100 x2 ⎛ ⎝⎜ ⎞ ⎠⎟ x2 2 − 5x ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→∞ 7+ 1x − 100 x2 2 − 5x = 72 28). lim x→∞ x2 − 3x + 7 x3 +10x − 4 = limx→∞ x3 1x − 3 x2 + 7 x3 ⎛ ⎝⎜ ⎞ ⎠⎟ x3 1+ 10x2 − 4 x3 ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→∞ 1 x − 3 x2 + 7 x3 1+ 10x2 − 4 x3 = 0 29). lim x→−∞ 7x2 + x +11 4 − x = limx→−∞ x2 7− 1x + 11 x2 ⎛ ⎝⎜ ⎞ ⎠⎟ x2 4x2 − 1 x ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→−∞ 7+ 1x + 11 x2 4 x2 − 1 x Note que: 4x2 − 1 x = 1 x 4 x −1 ⎛ ⎝⎜ ⎞ ⎠⎟ quando x→−∞ tanto 1 x quanto 4 x −1 tendem para valores negativos Portanto: lim x→−∞ 7+ 1x + 11 x2 4 x2 − 1 x = ∞ 30). lim x→∞ x3 + 7x 4x3 + 5 = limx→∞ x3 + 7x 4x3 + 5 = limx→∞ x3 1+ 7x2 ⎛ ⎝⎜ ⎞ ⎠⎟ x3 4 + 5x3 ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→∞ 1+ 7x2 4 + 5x3 = 14 = 1 2 31). lim x→∞ x − x2 + 7 = ∞−∞ = indeterminação. Para resolver a indeterminação, fazemos: = lim x→∞ (x − x2 + 7) (x + x 2 + 7) (x + x2 + 7) = lim x→∞ x2 − (x2 + 7)x + x2 + 7 = lim x→∞ −7 x + x2 + 7 = −7 ∞ +∞ = −7 ∞ = 0 32). lim x→−∞ x − x2 + 7 = −∞ −∞ = −∞ 33). lim x→∞ x + 3 9x2 + 5x = lim x→∞ (x + 3)2 9x2 + 5x = lim x→∞ (x + 3)2 9x2 + 5x = lim x→∞ x2 + 6x + 9 9x2 + 5x = limx→∞ x2 + 6x + 9 9x2 + 5x = limx→∞ x2 1+ 6x + 9 x2 ⎛ ⎝⎜ ⎞ ⎠⎟ x2 9 + 5x ⎛ ⎝⎜ ⎞ ⎠⎟ = lim x→∞ 1+ 6x + 9 x2 9 + 5x = lim x→∞ 1 9 = 1 9 = 1 3 34). lim x→−∞ x + 3 9x2 + 5x = lim x→−∞ − (x + 3)2 9x2 + 5x = − lim x→−∞ (x + 3)2 9x2 + 5x = − lim x→−∞ x2 + 6x + 9 9x2 + 5x = − limx→−∞ x2 1+ 6x + 9 x2 ⎛ ⎝⎜ ⎞ ⎠⎟ x2 9 + 5x ⎛ ⎝⎜ ⎞ ⎠⎟ = − lim x→−∞ 1+ 6x + 9 x2 9 + 5x = − lim x→−∞ 1 9 = − 1 9 = − 1 3 35). lim x→∞ log x 6 − 500 x6 + 500 ⎛ ⎝⎜ ⎞ ⎠⎟ = log lim x→∞ x6 − 500 x6 + 500 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = log lim x→∞ x6 1− 500x6 ⎛ ⎝⎜ ⎞ ⎠⎟ x6 1+ 500x6 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = log lim x→∞ 1− 500x6 1+ 500x6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = log(1) = 0