Đồ thị hàm số \(y = \frac{{...

C

- Hướng dẫn giải

TXĐ: \({\rm{D}} = R\backslash \left\{ { \pm 3} \right\}.\) Ta có:

\(\mathop {\lim }\limits_{x \to {3^ - }} y = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{x - 2}}{{{x^2} - 9}} = - \infty ;\,{\rm{ }}\,\mathop {\lim }\limits_{x \to {3^ + }} y = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 2}}{{{x^2} - 9}} = + \infty \Rightarrow x = 3\) là TCĐ

\(\mathop {\lim }\limits_{x \to - {3^ - }} y = \mathop {\lim }\limits_{x \to - {3^ - }} \frac{{x - 2}}{{{x^2} - 9}} = + \infty ;\,\,\mathop {\lim }\limits_{x \to - {3^ + }} y = \mathop {\lim }\limits_{x \to - {3^ + }} \frac{{x - 2}}{{{x^2} - 9}} = - \infty \Rightarrow x = - 3\) TCĐ

\(\mathop {\lim }\limits_{x \to - \infty } y = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x} - \frac{2}{{{x^2}}}}}{{1 - \frac{9}{{{x^2}}}}} = 0;\,\,\mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x} - \frac{2}{{{x^2}}}}}{{1 - \frac{9}{{{x^2}}}}} = 0 \Rightarrow y = 0\) là TCN.

Vậy đồ thị hàm số có đúng ba tiệm cận.