Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
13,\\
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\sqrt x + 1}}{{{x^2} + x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{{x\sqrt x + 1}}{{{x^2}}}}}{{\dfrac{{{x^2} + x + 1}}{{{x^2}}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{1}{{\sqrt x }} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}} = \dfrac{{0 + 0}}{{1 + 0 + 0}} = 0\\
15,\\
\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{3{x^3} - 2x + 2}}{{ - 2{x^3} + 2{x^2} - 1}} = \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\dfrac{{3{x^3} - 2x + 2}}{{{x^3}}}}}{{\dfrac{{ - 2{x^3} + 2{x^2} - 1}}{{{x^3}}}}}\\
= \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{3 - \dfrac{2}{{{x^2}}} + \dfrac{2}{{{x^3}}}}}{{ - 2 + \dfrac{2}{x} - \dfrac{1}{{{x^3}}}}} = \dfrac{{3 - 0 + 0}}{{ - 2 + 0 - 0}} = - \dfrac{3}{2}
\end{array}\)
