Đáp án:
$\begin{array}{l}
7)\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - 1 + 1 - \sqrt[3]{{1 + 3x}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2x}}{{\sqrt {1 + 2x} + 1}} + \frac{{ - 3x}}{{1 + \sqrt[3]{{1 + 3x}} + {{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2}}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{2}{x}}}{{\sqrt {1 + 2x} + 1}} - \frac{{\frac{3}{x}}}{{1 + \sqrt[3]{{1 + 3x}} + {{\left( {\sqrt[3]{{1 + 3x}}} \right)}^2}}}} \right)\\
= 0\\
10)\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x + {x^2} + 2{x^3}} - \sqrt[3]{{1 + 3x + 3{x^2}}}}}{{{x^3}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2x + {x^2} + 2{x^3}}}{{\sqrt {1 + 2x + {x^2} + 2{x^3}} + 1}} - \frac{{3x + 3{x^2}}}{{1 + \sqrt[3]{{1 + 3x + 3{x^2}}} + {{\left( {\sqrt[3]{{1 + 3x + 3{x^2}}}} \right)}^2}}}}}{{{x^3}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{2}{{{x^2}}} + \frac{1}{x} + 2}}{{\sqrt {1 + 2x + {x^2} + 2{x^3}} + 1}} - \frac{{\frac{3}{{{x^2}}} + 3}}{{1 + \sqrt[3]{{1 + 3x + 3{x^2}}} + {{\left( {\sqrt[3]{{1 + 3x + 3{x^2}}}} \right)}^2}}}\\
= \frac{2}{2} - \frac{3}{3} = 0
\end{array}$
