Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 3{x^2} + 2x + 2 - \sqrt {x + 2} }}{{{x^3} - 8}}\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{{x^3} - 3{x^2} + 2x}}{{{x^3} - 8}} + \frac{{2 - \sqrt {x + 2} }}{{{x^3} - 8}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{x\left( {{x^2} - 3x + 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} + \frac{{\left( {2 - \sqrt {x + 2} } \right)\left( {2 + \sqrt {x + 2} } \right)}}{{\left( {{x^3} - 8} \right)\left( {2 + \sqrt {x + 2} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} + \frac{{{2^2} - \left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\left( {2 + \sqrt {x + 2} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{x\left( {x - 1} \right)}}{{{x^2} + 2x + 4}} + \frac{{ - 1}}{{\left( {{x^2} + 2x + 4} \right)\left( {2 + \sqrt {x + 2} } \right)}}} \right]\\
 = \frac{{2.\left( {2 - 1} \right)}}{{{2^2} + 2.2 + 4}} + \frac{{ - 1}}{{\left( {{2^2} + 2.2 + 4} \right)\left( {2 + \sqrt {2 + 2} } \right)}}\\
 = \frac{7}{{48}}\\
*)\\
\mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to  - 1} \frac{{\sqrt[3]{{x + 2}} - 1}}{{{x^3} - x}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {\sqrt[3]{{x + 2}} - 1} \right)\left( {{{\sqrt[3]{{x + 2}}}^2} + \sqrt[3]{{x + 2}}.1 + {1^2}} \right)}}{{\left( {{x^3} - x} \right)\left( {{{\sqrt[3]{{x + 2}}}^2} + \sqrt[3]{{x + 2}}.1 + {1^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 2} \right) - {1^3}}}{{x\left( {{x^2} - 1} \right)\left( {{{\sqrt[3]{{x + 2}}}^2} + \sqrt[3]{{x + 2}} + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{x + 1}}{{x\left( {x - 1} \right)\left( {x + 1} \right)\left( {{{\sqrt[3]{{x + 2}}}^2} + \sqrt[3]{{x + 2}} + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{1}{{x\left( {x - 1} \right)\left( {{{\sqrt[3]{{x + 2}}}^2} + \sqrt[3]{{x + 2}} + 1} \right)}}\\
 = \frac{1}{{\left( { - 1} \right).\left( { - 1 - 1} \right)\left( {{{\sqrt[3]{{ - 1 + 2}}}^2} + \sqrt[3]{{ - 1 + 2}} + 1} \right)}}\\
 = \frac{1}{6}\\
f\left( { - 1} \right) =  - {\left( { - 1} \right)^2} - 3.\left( { - 1} \right) - \frac{{11}}{6} = \frac{1}{6}\\
 \Rightarrow \mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = f\left( { - 1} \right)
\end{array}\)

Do đó, hàm số đã cho liên tục tại \(x =  - 1\)

\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sqrt {{x^2} + 1}  - 1}}{{4 - \sqrt {{x^2} + 16} }}\\
 = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left( {\sqrt {{x^2} + 1}  - 1} \right)\left( {\sqrt {{x^2} + 1}  + 1} \right)\left( {4 + \sqrt {{x^2} + 16} } \right)}}{{\left( {4 - \sqrt {{x^2} + 16} } \right)\left( {4 + \sqrt {{x^2} + 16} } \right)\left( {\sqrt {{x^2} + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left( {\left( {{x^2} + 1} \right) - {1^2}} \right)\left( {4 + \sqrt {{x^2} + 16} } \right)}}{{\left( {{4^2} - \left( {{x^2} + 16} \right)} \right)\left( {\sqrt {{x^2} + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^2}\left( {4 + \sqrt {{x^2} + 16} } \right)}}{{ - {x^2}\left( {\sqrt {{x^2} + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{4 + \sqrt {{x^2} + 16} }}{{ - \left( {\sqrt {{x^2} + 1}  + 1} \right)}}\\
 = \frac{{4 + \sqrt {{0^2} + 16} }}{{ - \left( {\sqrt {{0^2} + 1}  + 1} \right)}} =  - 4\\
\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - 4{x^3} - x - 4} \right) =  - {4.0^3} - 0 - 4 =  - 4\\
 \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)
\end{array}\)

Do đó, hàm số đã cho liên tục tại \(x =  0\)