Cho hàm số f(x) = (3x^2 − 10x + 3)/(x^2 − 5x + 6). Xét tính đúng sai của các khẳng định sau:
Hướng dẫn giải:
Đáp án: a) Sai. b) Đúng. c) Sai. d) Sai.
a) Sai. \(f\left( x \right) = \frac{{3{x^2} - 10x + 3}}{{{x^2} - 5x + 6}} = \frac{{\left( {x - 3} \right)\left( {3x - 1} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\).
b) Đúng. Ta có: \[\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {3x - 1} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{3x - 1}}{{x - 2}} = 8\].
c) Sai. \[\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 1}}{{x - 2}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3 - \frac{1}{x}}}{{1 - \frac{2}{x}}} = 3\].
d) Sai. \[\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{3x - 1}}{{x - 2}} = + \infty \] vì \[\mathop {\lim }\limits_{x \to {2^ + }} \left( {3x - 1} \right) = 5 > 0;\,\,\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0;\,\,x - 2 > 0\]