Giải thích các bước giải:

\(\begin{array}{l}
b,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - x + 2} \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt {{x^2} + x}  - \left( {x - 2} \right)} \right]\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left[ {\frac{{\left( {{x^2} + x} \right) - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2} + x}  + x - 2}}} \right]\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{5x - 4}}{{\sqrt {{x^2} + x}  + x - 2}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{5 - \frac{4}{x}}}{{\sqrt {1 + \frac{1}{x}}  + 1 - \frac{2}{x}}}\\
 = \frac{5}{{\sqrt 1  + 1}}\\
 = \frac{5}{2}\\
d,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {2x + 1 - \sqrt {4{x^2} + 4x + 3} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{{\left( {2x + 1} \right)}^2} - \left( {4{x^2} + 4x + 3} \right)}}{{2x + 1 + \sqrt {4{x^2} + 4x + 3} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 2}}{{2x + 1 + \sqrt {4{x^2} + 4x + 3} }}\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {2x + 1 + \sqrt {4{x^2} + 4x + 3} } \right) =  + \infty \\
 \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 2}}{{2x + 1 + \sqrt {4{x^2} + 4x + 3} }} = 0\\
h,\\
\mathop {\lim }\limits_{x \to  + \infty } \sqrt {x - 2} .\left( {\sqrt {x + 3}  - \sqrt {x - 1} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {x - 2} .\frac{{\left( {x + 3} \right) - \left( {x - 1} \right)}}{{\sqrt {x + 3}  + \sqrt {x - 1} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {x - 2} .\frac{4}{{\sqrt {x + 3}  + \sqrt {x - 1} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4\sqrt {x - 2} }}{{\sqrt {x + 3}  + \sqrt {x - 1} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4\sqrt {1 - \frac{2}{x}} }}{{\sqrt {1 + \frac{3}{x}}  + \sqrt {1 - \frac{1}{x}} }}\\
 = \frac{{4\sqrt 1 }}{{\sqrt 1  + \sqrt 1 }}\\
 = 2\\
j,\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt[3]{{{x^3} - {x^2} + 3x}} - x + 1} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\sqrt[3]{{{x^3} - {x^2} + 3x}} - \left( {x - 1} \right)} \right]\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{{\sqrt[3]{{{x^3} - {x^2} + 3x}}}^3} - {{\left( {x - 1} \right)}^3}}}{{{{\sqrt[3]{{{x^3} - {x^2} + 3x}}}^2} + \left( {x - 1} \right)\sqrt[3]{{{x^3} - {x^2} + 3x}} + {{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^3} - {x^2} + 3x - {x^3} + 3{x^2} - 3x + 1}}{{{{\sqrt[3]{{{x^3} - {x^2} + 3x}}}^2} + \left( {x - 1} \right)\sqrt[3]{{{x^3} - {x^2} + 3x}} + {{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{2{x^2} + 1}}{{{{\sqrt[3]{{{x^3} - {x^2} + 3x}}}^2} + \left( {x - 1} \right)\sqrt[3]{{{x^3} - {x^2} + 3x}} + {{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{2 + \frac{1}{{{x^2}}}}}{{{{\sqrt[3]{{1 - \frac{1}{x} + \frac{3}{{{x^2}}}}}}^2} + \left( {1 - \frac{1}{x}} \right)\sqrt[3]{{1 - \frac{1}{x} + \frac{3}{{{x^2}}}}} + {{\left( {1 - \frac{1}{x}} \right)}^2}}}\\
 = \frac{2}{{1 + 1 + 1}}\\
 = \frac{2}{3}
\end{array}\) 

\(\begin{array}{l}
l,\\
\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^3} - 2{x^2} - x}} - x} \right)\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} - 2{x^2} - x - {x^3}}}{{{{\sqrt[3]{{{x^3} - 2{x^2} - x}}}^2} + x.\sqrt[3]{{{x^3} - 2{x^2} - x}} + {x^2}}}\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 2{x^2} - x}}{{{{\sqrt[3]{{{x^3} - 2{x^2} - x}}}^2} + x.\sqrt[3]{{{x^3} - 2{x^2} - x}} + {x^2}}}\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 2 - \frac{1}{x}}}{{{{\sqrt[3]{{1 - \frac{2}{x} - \frac{1}{{{x^2}}}}}}^2} + 1.\sqrt[3]{{1 - \frac{2}{x} - \frac{1}{{{x^2}}}}} + 1}}\\
 = \frac{{ - 2}}{3}
\end{array}\)