Đáp án:
\(2)\dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{1}{x}}}{{2 + \dfrac{3}{x}}} = \dfrac{0}{2} = 0\\
2)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 - \dfrac{1}{x}}}{{\dfrac{1}{x} + 3}} = \dfrac{2}{3}\\
3)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x + 1}}{{1 - {x^2}}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}}}}{{\dfrac{1}{{{x^2}}} - 1}} = - 2\\
4)\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{2 - x}}{{x - 1}} = - \infty \\
Do:x \to {1^ - } \to x - 1 < 0
\end{array}\)
