Đáp án:
$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to - \infty } \frac{{2x - 3}}{{3x + 7}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 - \frac{3}{x}}}{{3 + \frac{7}{x}}} = \frac{{2 - 0}}{{3 + 0}} = \frac{2}{3}\\
2)\mathop {\lim }\limits_{x \to + \infty } \frac{{2x + 9}}{{{x^3} + x - 5}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{2}{{{x^2}}} + \frac{9}{{{x^3}}}}}{{1 + \frac{1}{{{x^2}}} - \frac{5}{{{x^3}}}}} = 0\\
3)\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{{2{x^2} - x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {\frac{{{x^2} + 1}}{{{x^4}}}} }}{{2 - \frac{1}{x} + \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {\frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}} }}{{2 - \frac{1}{x} + \frac{1}{{{x^2}}}}} = 0
\end{array}$
