Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 6x} - \sqrt[3]{{1 + 9x}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {1 + 6x} - \left( {3x + 1} \right)} \right) + \left( {\left( {3x + 1} \right) - \sqrt[3]{{1 + 9x}}} \right)}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 6x} - \left( {3x + 1} \right)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {3x + 1} \right) - \sqrt[3]{{1 + 9x}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 6x - {{\left( {3x + 1} \right)}^2}}}{{\left( {\sqrt {1 + 6x} + \left( {3x + 1} \right)} \right){x^2}}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {3x + 1} \right)}^3} - \left( {1 + 9x} \right)}}{{\left( {{{\left( {3x + 1} \right)}^2} + \left( {3x + 1} \right)\sqrt[3]{{1 + 9x}} + {{\left( {\sqrt[3]{{1 + 9x}}} \right)}^2}} \right){x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 9{x^2}}}{{\left( {\sqrt {1 + 6x} + 3x + 1} \right){x^2}}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{27{x^3} + 27{x^2}}}{{\left( {{{\left( {3x + 1} \right)}^2} + \left( {3x + 1} \right)\sqrt[3]{{1 + 9x}} + {{\left( {\sqrt[3]{{1 + 9x}}} \right)}^2}} \right){x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 9}}{{\sqrt {1 + 6x} + 3x + 1}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{27x + 27}}{{{{\left( {3x + 1} \right)}^2} + \left( {3x + 1} \right)\sqrt[3]{{1 + 9x}} + {{\left( {\sqrt[3]{{1 + 9x}}} \right)}^2}}}\\
= \dfrac{{ - 9}}{{\sqrt {1 + 0} + 0 + 1}} + \dfrac{{0 + 27}}{{{1^2} + 1.1 + {1^2}}}\\
= \dfrac{9}{2}
\end{array}$
