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

\(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right]\)\( = \mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = - 3 + 5 = 2\).

a) Hàm số f(x) xác định trên ℝ.

b) \(f\left( { - 3} \right) = a - \frac{{11}}{9}\).

c) \(\mathop {\lim }\limits_{x \to - 3} f\left( x \right) = \mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} - 9}}{{{x^3} + 27}}\).

d) Có \(\mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} - 9}}{{{x^3} + 27}} = \mathop {\lim }\limits_{x \to - 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right)}}\)\( = \mathop {\lim }\limits_{x \to - 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right)}}\)\( = \mathop {\lim }\limits_{x \to - 3} \frac{{x - 3}}{{{x^2} - 3x + 9}} = - \frac{2}{9}\).

Hàm số gián đoạn tại x = −3 khi \(\mathop {\lim }\limits_{x \to - 3} f\left( x \right) \ne f\left( { - 3} \right)\)Û \(a - \frac{{11}}{9} \ne - \frac{2}{9}\)\( \Leftrightarrow a \ne 1\).

Mà a nguyên, a Î (0; 25) nên có 23 giá trị nguyên thỏa mãn.

Đáp án: a) Đúng;   b) Đúng;   c) Đúng;   d) Đúng.