Giải thích các bước giải:
\(\begin{array}{l}
s,\\
\mathop {\lim }\limits_{x \to - 2} \frac{{\sqrt {{x^3} + 8{x^2} - 8} + 2x}}{{\sqrt {{x^2} + 5} - \sqrt {2{x^2} + 1} }}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{\frac{{{x^3} + 8{x^2} - 8 - 4{x^2}}}{{\sqrt {{x^3} + 8{x^2} - 8} - 2x}}}}{{\frac{{{x^2} + 5 - 2{x^2} - 1}}{{\sqrt {{x^2} + 5} + \sqrt {2{x^2} + 1} }}}}\\
= \mathop {\lim }\limits_{x \to - 2} \left( {\frac{{{x^3} + 4{x^2} - 8}}{{\sqrt {{x^3} + 8{x^2} - 8} - 2x}}:\frac{{ - {x^2} + 4}}{{\sqrt {{x^2} + 5} + \sqrt {2{x^2} + 1} }}} \right)\\
= \mathop {\lim }\limits_{x \to - 2} \left( {\frac{{\left( {x + 2} \right)\left( {{x^2} + 2x - 4} \right)}}{{\sqrt {{x^3} + 8{x^2} - 8} - 2x}}:\frac{{\left( {2 - x} \right)\left( {2 + x} \right)}}{{\sqrt {{x^2} + 5} + \sqrt {2{x^2} + 1} }}} \right)\\
= \mathop {\lim }\limits_{x \to - 2} \left( {\frac{{{x^2} + 2x - 4}}{{\sqrt {{x^3} + 8{x^2} - 8} - 2x}}.\frac{{\sqrt {{x^2} + 5} + \sqrt {2{x^2} + 1} }}{{2 - x}}} \right)\\
= \frac{{{{\left( { - 2} \right)}^2} + 2.\left( { - 2} \right) - 4}}{{\sqrt {{{\left( { - 2} \right)}^3} + 8.{{\left( { - 2} \right)}^2} - 8} - 2.\left( { - 2} \right)}}.\frac{{\sqrt {{{\left( { - 2} \right)}^2} + 5} + \sqrt {2.{{\left( { - 2} \right)}^2} + 1} }}{{2 - \left( { - 2} \right)}}\\
= \frac{{ - 4}}{8}.\frac{6}{4} = - \frac{3}{4}
\end{array}\)
