Calculo
(c) lim (−x5 + 6x4 + 2)
(d) lim (2x + 7) x→1/2 (b) lim (3x2 − 7x + 2) x→3 (f ) lim [(x − 2)10 · (x + 4)] x→0 t+3 (h) lim x→2 t + 2 t2 + 5t + 6 (j) lim t→2 t+2 s+4 (l) lim s→1/2 2s 2 (n) lim (3x + 2) 3 x→4 x→7 √ √ 2x2 − x x x− 2 (o) lim (p) lim √ x→2 3x 3x − 4 x→ 2 (q) lim [2 sin x − cos x + cot x] (r) lim (ex + 4x) π x→ 2 x→4
(e) lim [(x + 4)3 · (x + 2)−1 ] x→−1 x+4 (g) lim x→2 3x − 1 x2 − 1 (i) lim x→1 x − 1 t2 − 5t + 6 (k) lim t→2 √t − 2 (m) lim 3 2x + 3
(s) lim1 (2x + 3) x→− 3
1 4
Exercício 2. Seja f (x) =
x − 1, x 3 3x − 7, x > 3. t→3 Calcule:
(a) lim f (x) (b) lim f (x) lim f (x) − + t→3 t→3
(d) lim f (x) (e) lim f (x) lim f (x). − + t→5 t→5 t→5
Esboçar o gráfico de f (x). Exercício 3. Seja F (x) = |x − 4|. Calcule os limites indicados se existirem: (a)a; lim+ F (x). (b) lim F (x). (c) lim F (x). − x→4 x→4 x→4
Esboce o gráfico de F (x). Exercício 4. Seja f (x) = 2 + |5x − 1|. Calcule se existir: (a) lim+ f (x). (b) lim f (x). (c) lim f (x). −1
1 x→ 5 1 x→ 5
x→1/5
1
Esboce o gráfico de f (x). x , se x = 0 Exercício 5. Seja h(x) = |x| Mostrar que h(x) não tem limite 0, se x = 0. no ponto 0. Exercício 6. Verifique se lim 1 existe. x→1 x − 1 x2 + 3x − 10 . x→2 3x2 − 5x − 2 x2 + 6x + 5 (f ) lim 2 . x→−1 x − 3x − 4 x2 − 5x + 6 (i) lim 2 . x→2 x − 12x = 20 √ 25 + 3t − 5 (l) lim t→0 √ t 3 8+h−2 (o) lim . h→0 √ h √ 1+x− 1−x (r) lim . x→0 x (c) lim
Exercício 7. Calcule os seguintes limites: x3 + 1 . x→−1 x2 − 1 2t2 − 3t − 5 (d) lim . t→5/2 2t − 5 x2 − 1 (g) lim 2 x→−1 x + 3x + 2 (2 + h)4 − 16 (j) lim h→0 √ h h−1 . (m) lim h→1√h − 1 1+x−1 (p) lim . x→0 −x (a) lim t3 + 4t2 + 4t t→−2 (t + 2)(t − 3) 3x2 − 17x + 20 (e) lim 2 . x→4 4x − 25x + 36 x2 − 4 (h) lim x→2 x − 2 (4 + t)2 − 16 (k) lim . t→0 t 2(h2 − 8) + h (n) lim . h→−4 √ h+4 3− 5+x √ . (q)