Trong bốn giới hạn sau đây, giới hạn nào bằng −∞?

65/100

\(\mathop {\lim }\limits_{x \to + \infty } \frac{{ - 3x + 4}}{{x - 2}}\).

\(\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3x + 4}}{{x - 2}}\).

\(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{ - 3x + 4}}{{x - 2}}\).

\(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{ - 3x + 4}}{{x - 2}}\).

Giải thích

Dễ thấy \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ - 3x + 4}}{{x - 2}} = - 3;\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3x + 4}}{{x - 2}} = - 3\) (loại).

Vì \(\mathop {\lim }\limits_{x \to {2^ - }} ( - 3x + 4) = - 2 < 0;\mathop {\lim }\limits_{x \to {2^ - }} (x - 2) = 0;x - 2 < 0,\forall x < 2{\rm{ n\^e n }}\mathop {\lim }\limits_{x \to {2^ - }} \frac{{ - 3x + 4}}{{x - 2}} = + \infty {\rm{.}}\)

Vì \(\mathop {\lim }\limits_{x \to {2^ + }} ( - 3x + 4) = - 2 < 0;\mathop {\lim }\limits_{x \to {2^ + }} (x - 2) = 0;x - 2 > 0,\forall x > 2{\rm{ n\^e n }}\mathop {\lim }\limits_{x \to {2^ + }} \frac{{ - 3x + 4}}{{x - 2}} = - \infty \).