PHẦN II. TRẮC NGHIỆM ĐÚNG – SAI

Cho hai hàm số y = f(x); y = g(x) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5\) và \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = + \infty \).

a) \(\mathop {\lim }\limits_{x \to 2} \left[ {5f\left( x \right)} \right] = - \infty \).

b) \(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right).g\left( x \right)} \right] = + \infty \).

c) \[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = + \infty \].

d) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) - 1} - 2}}{{f\left( x \right) - 5}} = \frac{1}{4}\).

\(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{x + 2}}{{x - 1}} =  - \infty \) vì \[\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {x + 2} \right) = 3 > 0\\\mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 0\\x - 1 < 0,\forall x < 1\end{array} \right.\].

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

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

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^3} - {x^2} + x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} + 1}} =  - \frac{1}{2} < 0\).

d) Để \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = 2\) thì ax + b có nghiệm bằng 1 Û a + b = 0 Û b = −a.

Khi đó \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{a\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{a} =  - \frac{1}{a} = 2\) \( \Leftrightarrow a =  - \frac{1}{2} \Rightarrow b = \frac{1}{2}\).

Suy ra \(a + 3b =  - \frac{1}{2} + 3.\frac{1}{2} = 1\).

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