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Question

Tìm các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 5x - 2} \right)$;

b) $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}}$.

Quoc Tran Anh Le · Accepted Answer

a) $\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 5x - 2} \right) = \mathop {\lim }\limits_{x \to - 2} {x^2} + \mathop {\lim }\limits_{x \to - 2} \left( {5x} \right) - \mathop {\lim }\limits_{x \to - 2} 2$

$ = \mathop {\lim }\limits_{x \to - 2} {x^2} + 5\mathop {\lim }\limits_{x \to - 2} x - \mathop {\lim }\limits_{x \to - 2} 2 = {\left( { - 2} \right)^2} + 5.\left( { - 2} \right) - 2 = - 8$

b) $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2$

Câu trả lời:

a) $\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 5x - 2} \right) = \mathop {\lim }\limits_{x \to - 2} {x^2} + \mathop {\lim }\limits_{x \to - 2} \left( {5x} \right) - \mathop {\lim }\limits_{x \to - 2} 2$

$ = \mathop {\lim }\limits_{x \to - 2} {x^2} + 5\mathop {\lim }\limits_{x \to - 2} x - \mathop {\lim }\limits_{x \to - 2} 2 = {\left( { - 2} \right)^2} + 5.\left( { - 2} \right) - 2 = - 8$

b) $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2$

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