... 3) n n + 2n − 3n n3 ( a) 0,25 x − 11 x0 − = ⇔ x0 = 11 x0 + ≥ 11 x0 Đề 1: Câu 1( 7,0đ ) 0,25 3 x0 ≥ 11 x0 ⇔ x0 ≥ 44 x0 ⇔ x0 ≥ 44 0,25 x0 ≥ 44 0,25 Điể m 1, 0 2,0 đ 1, 0 + − 9) = lim n n ( + − 3) =3 n ... đ x 1 x 1 0,5 1, 0 x2 + x − x − = +∞ lim ( x − 2) x →+∞ c) 1, 0 x +1 ( x − 2) ( x + 1) = lim x3 + x − x →+∞ x + x − 1, 0 2 x (1 − )2 (1 + ) (1 − ) (1 + ) x x = lim x x x →+∞ 1 x (1 − ) (1 − ) ... = 1 b) Ta có: x →−2 x + x → 0− x − 2x + x2 + 2x x → −2− 1, 0đ 1, 0đ lim− Vậy 2,0 đ x →−2 lim ( x + 5) x →+∞ c) = lim x →+∞ 1, 0đ = +∞ 1, 0đ 5 x (1 + ) 1+ x = lim x x →+∞ 1 x (1 − ) 1 x x =1 lim...