Limite- calculo1

UNIVERSIDADE FEDERAL DE PERNAMBUCO ´ CCEN–DEPARTAMENTO DE MATEMATICA C´lculo 1 - 1 a Lista de Exerc´ a ıcios - Solu¸˜es co
Prof. C´sar Castilho e Enviar corre¸˜es para [e-mail] co

1 - Calcule os seguintes limites: a) lim x2 − 4 x = −4 . x→2 b) lim x3 + 2 x2 − 3 x − 4 = 0 . x→−1 c) lim

(3 x − 1)2 1 = . 3 x→1 (x + 1) 2 3x − 3−x 30 − 30 1−1 = 0 = = 0. x + 3−x 0 x→0 3 3 +3 1+1 .

d) lim e)
1 3

f) lim

x2 − 4 (x − 2) (x + 2) (x + 2) 4 = lim = lim = . 2 − 5x + 6 x→2 x x→2 (x − 2) (x + 3) x→2 (x + 3) 5 x2 + 3 x + 2 (x + 1) (x + 2) (x + 2) 1 = lim = lim = 2 + 4x + 3 x→−1 x x→−1 (x + 1) (x + 3) x→−1 (x + 3) 2

g) lim h) lim

x−2 1 1 x−2 = lim = lim = . 2−4 x→2 (x − 2) (x + 2) x→2 (x + 2) x→2 x 4 √ √ √ x−2 x−2 x−2 x−2 √ i) lim √ 2 = lim √ = lim √ = 0. x→2 x − 4 x→2 x − 2 x + 2 x→2 x + 2 √ √ x−2 x−2 1 √ j) lim 2 = lim √ = lim √ = +∞ . x→2+ x − 4 x→2+ x − 2 x − 2 (x + 2) x→2+ x − 2 (x + 2) k) lim x3 + 3 x2 h + 3 x h 2 + h 3 − x3 h (3 x2 + 3 x h + h2 ) (x + h)3 − x3 = lim = lim = h→0 h→0 h→0 h h h

= lim (3 x2 + 3 x h + h2 ) = 3 x2 . h→0 √ √ x−1 (x − 1) ( x2 + 3 + 2) (x − 1) ( x2 + 3 + 2) √ l) lim √ 2 = lim √ = lim = x→1 (x2 − 1) x + 3 − 2 x→1 ( x2 + 3 − 2) ( x2 + 3 + 2) x→1

√ ( x2 + 3 + 2) = lim = 2. x→1 (x + 1)

2 - Calcule os seguintes limites: 3 x2 − 2 x 1) lim = x→3 x + 1 4 2) lim x→0 x3

6x − 9 = −3 − 12 x + 3

3) lim 4) lim 5) lim 6) lim 7)lim

(t + 2) (t2 − 2 t + 4) t3 + 8 = lim = lim (t2 − 2 t + 4) = 12 . t→−2 t→−2 t→−2 t + 2 t+2

t3 + t2 − 5 t + 3 4 = . 2 − 3t + 2 t→−1 t 3

(x − 4) (x + 4) x2 − 16 = lim = 8. x→4 x→4 x − 4 x−4 x2 − 4 x + 4 (x − 2)2 (x − 2) = lim = lim = 0. x→2 x2 + x − 6 x→2 (x + 3) (x − 2) x→2 (x + 3)

t3 + t2 − 5 t + 3 (t − 1)2 (t + 3) (t + 3) 4 = lim = lim = . 3 − 3t + 2 2 (t + 2) t→1 t→1 (t − 1) t→1 (t + 2) t 3 x→3 8) lim + 9) lim + x→1 x = +∞ . x−3

x4 − 1 (x − 1) (x3 + x2 + x + 1) = lim = lim (x3 + x2 + x + 1) = 4 . x→1+ x→1+ x−1 x−1 y+6 1 = . 2 + 36 y 6 x2 3−x 3−x = lim = +∞