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