Tính các giới hạn của các hàm số sau:\(A = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x - 6}}{{{x^2} - 3x + 2}}\)\(B = \mathop {\

\(A = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x - 6}}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x + 3}}{{x - 1}} = 5\)

\(B = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 4x + 5}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{{x^2} + 4x + 5 - {x^2}}}{{\sqrt {{x^2} + 4x + 5}  + x}}} \right) = \mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{4 + \frac{5}{x}}}{{\sqrt {1 + \frac{4}{x} + \frac{5}{{{x^2}}}}  + 1}}} \right) = 2\)