Question

Tìm các giới hạn sau:

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

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

Answer 1

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