Đáp án:

h) \(\dfrac{{11}}{2}\)

Giải thích các bước giải:

\(\begin{array}{l}
g)\mathop {\lim }\limits_{x \to  - \infty } {x^3}\left( {1 + \dfrac{5}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right) =  - \infty \\
Do:\mathop {\lim }\limits_{x \to  - \infty } {x^3} =  - \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \dfrac{5}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right) = 1\\
h)\lim \dfrac{{9 + \sqrt {4 + \dfrac{1}{{{n^2}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{n^3}}}}} + 1}} = \dfrac{{9 + 2}}{{1 + 1}} = \dfrac{{11}}{2}\\
i)\mathop {\lim }\limits_{x \to  + \infty } x.\dfrac{{{x^2} + 5x + \dfrac{1}{x}}}{{ - {x^2} + x + 1}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } x.\dfrac{{1 + \dfrac{5}{x} + \dfrac{1}{{{x^3}}}}}{{ - 1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}} =  - \infty \\
Do:\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\
\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{1 + \dfrac{5}{x} + \dfrac{1}{{{x^3}}}}}{{ - 1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}} =  - 1\\
j)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3 - 5x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {\sqrt {x + 3}  + \sqrt {5x - 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {\sqrt {x + 3}  + \sqrt {5x - 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4}}{{\left( {{x^2} + x + 1} \right)\left( {\sqrt {x + 3}  + \sqrt {5x - 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 4}}{{\left( {{1^2} + 1 + 1} \right)\left( {\sqrt {1 + 3}  + \sqrt {5.1 - 1} } \right)}} = \dfrac{{ - 4}}{{3.4}} =  - \dfrac{1}{3}\\
k)\mathop {\lim }\limits_{x \to  - \infty } \left( {\dfrac{{{x^2} + 4x - {x^2} - x - 1}}{{\sqrt {{x^2} + 4x}  + \sqrt {{x^2} + x + 1} }}} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{3x - 1}}{{\sqrt {{x^2} + 4x}  + \sqrt {{x^2} + x + 1} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{3 - \dfrac{1}{x}}}{{ - \sqrt {1 + \dfrac{4}{x}}  - \sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }} =  - \dfrac{3}{2}\\
l)\mathop {\lim }\limits_{x \to 1} \dfrac{{ - 3{x^2} + 3x - 1 + 2x - 1}}{{\left( {x - 1} \right)\left( {\sqrt[3]{{{{\left( { - 3{x^2} + 3x - 1} \right)}^2}}} - \left( {2x - 1} \right)\sqrt[3]{{ - 3{x^2} + 3x - 1}} + {{\left( {2x - 1} \right)}^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {2 - 3x} \right)}}{{\left( {x - 1} \right)\left( {\sqrt[3]{{{{\left( { - 3{x^2} + 3x - 1} \right)}^2}}} - \left( {2x - 1} \right)\sqrt[3]{{ - 3{x^2} + 3x - 1}} + {{\left( {2x - 1} \right)}^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - 3x}}{{\sqrt[3]{{{{\left( { - 3{x^2} + 3x - 1} \right)}^2}}} - \left( {2x - 1} \right)\sqrt[3]{{ - 3{x^2} + 3x - 1}} + {{\left( {2x - 1} \right)}^2}}}\\
 = \dfrac{{2 - 3.1}}{{\sqrt[3]{{{{\left( { - {{3.1}^2} + 3.1 - 1} \right)}^2}}} - \left( {2.1 - 1} \right)\sqrt[3]{{ - {{3.1}^2} + 3.1 - 1}} + {{\left( {2.1 - 1} \right)}^2}}} = \dfrac{{ - 1}}{{1 + 1 + 1}} =  - \dfrac{1}{3}
\end{array}\)

( câu l mẫu số không có bình phương thì bài mới giải đc nha )