Giải thích các bước giải:

\(\begin{array}{l}
2.\\
a,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1}  - \sqrt {{x^2} + x + 1} }}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 1}  - \sqrt {{x^2} + x + 1} } \right)\left( {\sqrt {x + 1}  + \sqrt {{x^2} + x + 1} } \right)}}{{x.\left( {\sqrt {x + 1}  + \sqrt {{x^2} + x + 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {x + 1} \right) - \left( {{x^2} + x + 1} \right)}}{{x\left( {\sqrt {x + 1}  + \sqrt {{x^2} + x + 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{ - {x^2}}}{{x\left( {\sqrt {x + 1}  + \sqrt {{x^2} + x + 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{\sqrt {x + 1}  + \sqrt {{x^2} + x + 1} }}\\
 = \frac{{ - 0}}{{\sqrt {0 + 1}  + \sqrt {{0^2} + 0 + 1} }} = 0\\
b,\\
\mathop {\lim }\limits_{x \to 2} \frac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1}  - 3}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - \sqrt {x + 2} } \right)\left( {x + \sqrt {x + 2} } \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{\left( {\sqrt {4x + 1}  - 3} \right)\left( {\sqrt {4x + 1}  + 3} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} - \left( {x + 2} \right)} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{\left( {\left( {4x + 1} \right) - {3^2}} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} - x - 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{\left( {4x - 8} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x + 1} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4\left( {x - 2} \right)\left( {x + \sqrt {x + 2} } \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{9}{8}
\end{array}\)

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