Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
17,\\
x \to - \infty \Rightarrow x < 0 \Rightarrow x = - \sqrt {{x^2}} \\
\mathop {\lim }\limits_{x \to - \infty } x.\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} \\
= \mathop {\lim }\limits_{x \to - \infty } \left[ { - \sqrt {{x^2}} .\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} } \right]\\
= \mathop {\lim }\limits_{x \to - \infty } - \sqrt {\frac{{2{x^5} + {x^3}}}{{{x^5} - {x^2} + 3}}} \\
= - \mathop {\lim }\limits_{x \to - \infty } \sqrt {\frac{{2 + \frac{1}{{{x^2}}}}}{{1 - \frac{1}{{{x^3}}} + \frac{3}{{{x^5}}}}}} \\
= - \sqrt {\frac{2}{1}} = - \sqrt 2 \\
16,\\
x \to - \infty \Rightarrow x < 0 \Leftrightarrow \left\{ \begin{array}{l}
\left| x \right| = - x\\
\left| {3x + 2} \right| = - 3x - 2
\end{array} \right.\\
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {2{x^2} - 3x + 4} }}{{\left| {3x + 2} \right| + 4x - 7}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|.\sqrt {2 - \frac{3}{x} + \frac{4}{{{x^2}}}} }}{{ - 3x - 2 + 4x - 7}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x.\sqrt {2 - \frac{3}{x} + \frac{4}{{{x^2}}}} }}{{x - 9}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {2 - \frac{3}{x} + \frac{4}{{{x^2}}}} }}{{1 - \frac{9}{x}}}\\
= \frac{{ - \sqrt 2 }}{1}\\
= - \sqrt 2
\end{array}\)
