Giải thích các bước giải:

\(\begin{array}{l}
m,\\
\mathop {\lim }\limits_{x \to 2} \frac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1}  - 3}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{{x^2} - \left( {x + 2} \right)}}{{x + \sqrt {x + 2} }}}}{{\frac{{4x + 1 - {3^2}}}{{\sqrt {4x + 1}  + 3}}}}\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^2} - x - 2}}{{x + \sqrt {x + 2} }}:\frac{{4x - 8}}{{\sqrt {4x + 1}  + 3}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{x + \sqrt {x + 2} }}.\frac{{\sqrt {4x + 1}  + 3}}{{4\left( {x - 2} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x + 1} \right).\left( {\sqrt {4x + 1}  + 3} \right)}}{{4.\left( {x + \sqrt {x + 2} } \right)}}\\
 = \frac{{\left( {2 + 1} \right).\left( {\sqrt {4.2 + 1}  + 3} \right)}}{{4.\left( {2 + \sqrt {2 + 2} } \right)}}\\
 = \frac{{3.6}}{{4.4}} = \frac{9}{8}\\
n,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {\sqrt[3]{x} - 1} \right)}^2}}}{{{{\left[ {\left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)} \right]}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}^2}}}\\
 = \frac{1}{{{{\left( {\sqrt[3]{{{1^2}}} + \sqrt[3]{1} + 1} \right)}^2}}}\\
 = \frac{1}{9}\\
p,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 2x + 6}  - 4x + 1}}{{{x^3} - 2x + 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^2} + 2x + 6}  - 3} \right) - \left( {4x - 4} \right)}}{{{x^3} - {x^2} + {x^2} - x - x + 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{{x^2} + 2x + 6 - 9}}{{\sqrt {{x^2} + 2x + 6}  + 3}} - 4\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\sqrt {{x^2} + 2x + 6}  + 3}} - 4\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{1}{{{x^2} + x - 1}}.\left( {\frac{{x + 3}}{{\sqrt {{x^2} + 2x + 6}  + 3}} - 4} \right)} \right]\\
 = \frac{1}{{{1^2} + 1 - 1}}.\left( {\frac{{1 + 3}}{{\sqrt {{1^2} + 2.1 + 6}  + 3}} - 4} \right)\\
 = 1.\left( {\frac{4}{6} - 4} \right)\\
 =  - \frac{{10}}{3}\\
q,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[4]{x} - 1}}{{{x^3} + {x^2} - 2}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sqrt x  - 1}}{{\sqrt[4]{x} + 1}}.\frac{1}{{{x^3} + {x^2} - 2}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x - 1}}{{\left( {\sqrt x  + 1} \right).\left( {\sqrt[4]{x} + 1} \right)}}.\frac{1}{{{x^3} - {x^2} + 2{x^2} - 2x + 2x - 2}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x - 1}}{{\left( {\sqrt x  + 1} \right).\left( {\sqrt[4]{x} + 1} \right).\left( {x - 1} \right).\left( {{x^2} + 2x + 2} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {\sqrt x  + 1} \right).\left( {\sqrt[4]{x} + 1} \right).\left( {{x^2} + 2x + 2} \right)}}\\
 = \frac{1}{{2.2.5}}\\
 = \frac{1}{{20}}
\end{array}\)