Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
16,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{x + 7}} - \sqrt {5 - {x^2}} }}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt[3]{{x + 7}} - 2} \right) + \left( {2 - \sqrt {5 - {x^2}} } \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{x + 7 - {2^3}}}{{{{\sqrt[3]{{x + 7}}}^2} + 2.\sqrt[3]{{x + 7}} + 4}} + \frac{{{2^2} - 5 + {x^2}}}{{2 + \sqrt {5 - {x^2}} }}}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{x - 1}}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}} + \frac{{{x^2} - 1}}{{2 + \sqrt {5 - {x^2}} }}}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}} + \frac{{x + 1}}{{2 + \sqrt {5 - {x^2}} }}} \right)\\
 = \frac{1}{{{{\sqrt[3]{{1 + 7}}}^2} + 2.\sqrt[3]{{1 + 7}} + 4}} + \frac{{1 + 1}}{{2 + \sqrt {5 - {1^2}} }}\\
 = \frac{1}{{12}} + \frac{2}{4} = \frac{7}{{12}}\\
17,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3x + 4}  - \sqrt[3]{{8 + 5x}}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {3x + 4}  - 2} \right) + \left( {2 - \sqrt[3]{{8 + 5x}}} \right)}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{3x + 4 - {2^2}}}{{\sqrt {3x + 4}  + 2}} + \frac{{{2^3} - 8 - 5x}}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{3x}}{{\sqrt {3x + 4}  + 2}} - \frac{{5x}}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{3}{{\sqrt {3x + 4}  + 2}} - \frac{5}{{4 + 2\sqrt[3]{{8 + 5x}} + {{\sqrt[3]{{8 + 5x}}}^2}}}} \right]\\
 = \frac{3}{{\sqrt {3.0 + 4}  + 2}} - \frac{5}{{4 + 2.\sqrt[3]{{8 + 5.0}} + {{\sqrt[3]{{8 + 5.0}}}^2}}}\\
 = \frac{3}{4} - \frac{5}{{12}} = \frac{1}{3}
\end{array}\)