Tính các giới hạn sau: \(a/I = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{x - 2}}\), \[b/J = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - 5x + 3}}{{7{x^2} + x}}\]

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

b) \[J = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - x + 9}}{{7{x^2} + x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}} + \frac{9}{{{x^2}}}}}{{\frac{{7{x^2}}}{{{x^2}}} + \frac{x}{{{x^2}}}}}\]\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 - \frac{1}{x} + \frac{9}{{{x^2}}}}}{{7 + \frac{1}{x}}}\]\[ = \frac{1}{7}\]