Cho hàm số f(x) = (x^2 − 4x + 3)/(1 − x). Xét tính đúng/sai của các khẳng định sau:

Hướng dẫn giải:

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

a) Đúng. Ta có: \[\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 3}}{{1 - x}} = \frac{{{2^2} - 4 \cdot 2 + 3}}{{1 - 2}} = 1\].

b) Sai. Ta có: \[\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{1 - x}} = \frac{{{3^2} - 4 \cdot 3 + 3}}{{1 - 3}} = 0\].

c) Đúng. Ta có: \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{1 - x}} = \mathop {\lim }\limits_{x \to 1} \left( {3 - x} \right) = 3 - 1 = 2\].

d) Sai. Ta có:

\[\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - 4x + 3}}{{1 - x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {1 - \frac{4}{x} + \frac{3}{{{x^2}}}} \right)}}{{x\left( {\frac{1}{x} - 1} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {1 - \frac{4}{x} + \frac{3}{{{x^2}}}} \right)}}{{\left( {\frac{1}{x} - 1} \right)}} = - \infty \]