Ta có:
\(\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } {{1 - {x^2}} \over x} = \lim {{{x^2}({1 \over {{x^2}}} - 1)} \over {{x^2}.{1 \over x}}} = \lim {{{1 \over {{x^2}}} - 1} \over {{1 \over x}}}\)
Vì \(\mathop {\lim }\limits_{x \to - \infty } \left[ {{1 \over {{x^2}}} - 1} \right] = - 1 < 0\) (1)
\(\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x} = 0,\,\,x \to - \infty \Rightarrow \frac{1}{x} < 0\) (2)
Từ (1) và (2) suy ra: \(\mathop {\lim }\limits_{x \to - \infty } f(x)= +∞\)
Chọn đáp án A.
