Tính được các giới hạn sau: a) lim x → 2 ( 3 x ^2 − 2 x ) = 4

14/22

Tính được các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} - 2x} \right) = 4\)

b) \(\mathop {\lim }\limits_{x \to - 2} \frac{{4{x^2} + 2x + 1}}{{x - 4}} = \frac{{13}}{6}\)

c) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - {x^2} + 2x - 24}}{{{x^2} - 9}} = \frac{{23}}{6}\)

d) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} - x - 14}}{{{x^2} - 7x - 18}} = \frac{9}{{11}}\)

Nhập câu trả lời ở đây

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Giải thích

a) Sai

b) Sai

c) Đúng

d) Đúng

a) \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} - 2x} \right) = {3.2^2} - 2.2 = 8\)

b) \(\mathop {\lim }\limits_{x \to - 2} \frac{{4{x^2} + 2x + 1}}{{x - 4}} = - \frac{{13}}{6}\)

c) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^3} - {x^2} + 2x - 24}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{(x - 3)\left( {{x^2} + 2x + 8} \right)}}{{(x - 3)(x + 3)}} = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} + 2x + 8}}{{x + 3}} = \frac{{23}}{6}\)

d) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} - x - 14}}{{{x^2} - 7x - 18}} = \mathop {\lim }\limits_{x \to - 2} \frac{{(x + 2)\left( {{x^2} + 3x - 7} \right)}}{{(x + 2)(x - 9)}} = \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + 3x - 7}}{{x - 9}} = \frac{9}{{11}}\)