Tínha) \(A = \mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} + {x^2} - 5x - 6}}{{2{x^2} + 5x + 2}}\)b) \(B = \mathop {\lim }\limi

* Hướng dẫn giải

a) \(A = \mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} + {x^2} - 5x - 6}}{{2{x^2} + 5x + 2}}\)

\( = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x + 2} \right)\left( {{x^2} - x - 3} \right)}}{{\left( {x + 2} \right)\left( {2x + 1} \right)}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} - x - 3}}{{2x + 1}} =  - 1\)

b) \(B = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {25{x^2} + 10x}  - 5x} \right)\)

\( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {25{x^2} + 10x} \right) - 25{x^2}}}{{\sqrt {25{x^2} + 10x}  + 5x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{10}}{{\sqrt {25 + \frac{{10}}{x}}  + 5}} = 1\)

c) \(C = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \frac{{\sqrt {{x^2} - 4} }}{{{x^2} + 2x}}\)

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