Đáp án:
3) 0
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{1}{x} + 3}}{{\sqrt {2 + \dfrac{3}{{{x^2}}}} }}\\
= \dfrac{3}{{\sqrt 2 }}\\
2)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} - 1}}{{3 - {x^2}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 - \dfrac{1}{{{x^2}}}}}{{\dfrac{3}{{{x^2}}} - 1}}\\
= \dfrac{2}{{ - 1}} = - 2\\
3)\mathop {\lim }\limits_{x \to + \infty } \dfrac{5}{{3x + 2}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{5}{x}}}{{3 + \dfrac{2}{x}}} = \dfrac{0}{3} = 0
\end{array}\)
