Giải thích các bước giải:
\(\begin{array}{l}
2,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 3} - \sqrt[3]{{{x^3} + 7}}}}{{{x^3} - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^2} + 3} - 2} \right) + \left( {2 - \sqrt[3]{{{x^3} + 7}}} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{{x^2} + 3 - {2^2}}}{{\sqrt {{x^2} + 3} + 2}} + \frac{{{2^3} - {x^3} - 7}}{{4 + 2\sqrt[3]{{{x^3} + 7}} + {{\sqrt[3]{{{x^3} + 7}}}^2}}}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\sqrt {{x^2} + 3} + 2}} - \frac{{{x^3} - 1}}{{4 + 2\sqrt[3]{{{x^3} + 7}} + {{\sqrt[3]{{{x^3} + 7}}}^2}}}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x + 1}}{{\left( {{x^2} + x + 1} \right)\left( {\sqrt {{x^2} + 3} + 2} \right)}} - \frac{1}{{4 + 2\sqrt[3]{{{x^3} + 7}} + {{\sqrt[3]{{{x^3} + 7}}}^2}}}} \right]\\
= \frac{{1 + 1}}{{\left( {{1^2} + 1 + 1} \right).\left( {\sqrt {{1^2} + 3} + 2} \right)}} - \frac{1}{{4 + 2.\sqrt[3]{{{1^3} + 7}} + {{\sqrt[3]{{{1^3} + 7}}}^2}}}\\
= \frac{2}{{3.4}} - \frac{1}{{3.4}} = \frac{1}{{12}}\\
4,\\
\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{\sqrt[3]{{6 + 2x}} - \sqrt {1 + 3x} }}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {\sqrt[3]{{6 + 2x}} - 2} \right) + \left( {2 - \sqrt {1 + 3x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\frac{{6 + 2x - {2^3}}}{{{{\sqrt[3]{{6 + 2x}}}^2} + 2\sqrt[3]{{6 + 2x}} + 4}} + \frac{{4 - 1 - 3x}}{{2 + \sqrt {1 + 3x} }}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\frac{{2\left( {x - 1} \right)}}{{{{\sqrt[3]{{6 + 2x}}}^2} + 2\sqrt[3]{{6 + 2x}} + 4}} - \frac{{3\left( {x - 1} \right)}}{{2 + \sqrt {1 + 3x} }}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{\frac{2}{{{{\sqrt[3]{{6 + 2x}}}^2} + 2.\sqrt[3]{{6 + 2x}} + 4}} - \frac{3}{{2 + \sqrt {1 + 3x} }}}}\\
= \frac{{1 + 1}}{{\frac{2}{{{{\sqrt[3]{{6 + 2.1}}}^2} + 2.\sqrt[3]{{6 + 2.1}} + 4}} - \frac{3}{{2 + \sqrt {1 + 3.1} }}}}\\
= \frac{2}{{\frac{2}{{12}} - \frac{3}{4}}}\\
= - \frac{{24}}{7}
\end{array}\)
