Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 3x} + \sqrt[3]{{1 - 9x}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 3x} - 2 + 2 + \sqrt[3]{{1 - 9x}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 + 3x - 4}}{{\sqrt {1 + 3x} + 2}} + \frac{{8 + 1 - 9x}}{{4 - 2.\sqrt[3]{{1 - 9x}} + \sqrt[3]{{\left( {1 - 9{x^2}} \right)}}}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{3}{{\sqrt {1 + 3x} + 2}} + \frac{{ - 9}}{{4 - 2.\sqrt[3]{{1 - 9x}} + \sqrt[3]{{\left( {1 - 9{x^2}} \right)}}}}}}{{\left( {x - 1} \right)}}\\
= \frac{{\frac{3}{3} - \frac{9}{{4 - 2 + 1}}}}{0}\\
= \frac{{ - 2}}{0}\\
= - \infty
\end{array}$
