詳解：分母 \(\sqrt {{x^2} + 1} \) 恆正，故沒有鉛直漸近線

\(\lim\limits_{x \to \infty } f(x) = \lim\limits_{x \to \infty } \frac{{3x}}{{\sqrt {{x^2} + 1} }} = \lim\limits_{x \to \infty } \frac{3}{{\sqrt {1 + {\textstyle{1 \over {{x^2}}}}} }} = 3\)

\(\lim\limits_{x \to  - \infty } f(x) = \lim\limits_{x \to  - \infty } \frac{{3x}}{{\sqrt {{x^2} + 1} }} = \lim\limits_{x \to \infty } \frac{3}{{ - \sqrt {1 + {\textstyle{1 \over {{x^2}}}}} }} =  - 3\)

故有 \(y = 3\) 與 \(y =  - 3\) 兩條水平漸近線

函數概圖如下