Đáp án:
m) \( - \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
e)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt[3]{{3 + \dfrac{1}{{{x^3}}}}} + \sqrt {2 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }}{{ - \sqrt[4]{{4 + \dfrac{2}{{{x^4}}}}}}}\\
= \dfrac{{ - \sqrt[3]{3} + \sqrt 2 }}{{ - \sqrt[4]{4}}}\\
h)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2} - {x^2} - x - 1}}{{x - \sqrt {{x^2} + x + 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x - 1}}{{x - \sqrt {{x^2} + x + 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1 - \dfrac{1}{x}}}{{1 + \sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }} = \dfrac{{ - 1}}{2}\\
m)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {4{x^2} - 3x + 4 - 4{x^2}} \right)\left( {\sqrt {{x^2} + x + 1} - x} \right)}}{{\left( {{x^2} + x + 1 - {x^2}} \right)\left( {\sqrt {4{x^2} - 3x + 4} - 2x} \right)}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( { - 3x + 4} \right)\left( {\sqrt {{x^2} + x + 1} - x} \right)}}{{\left( {x + 1} \right)\left( {\sqrt {4{x^2} - 3x + 4} - 2x} \right)}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( { - 3 + \dfrac{4}{x}} \right)\left( { - \sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} - 1} \right)}}{{\left( {1 + \dfrac{1}{x}} \right)\left( { - \sqrt {4 - \dfrac{3}{x} + \dfrac{4}{{{x^2}}}} - 2} \right)}}\\
= \dfrac{{ - 3.\left( { - 2} \right)}}{{1.\left( { - 4} \right)}} = - \dfrac{3}{2}
\end{array}\)
