Đáp án:
$5$
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^6} - 1}}{{x\left( {{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^5} + {{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^4} + ... + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + 4x} \right)}^3}{{\left( {1 + 9x} \right)}^2} - 1}}{{x\left( {{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^5} + {{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^4} + ... + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {64{x^3} + 48{x^2} + 12x + 1} \right)\left( {81{x^2} + 18x + 1} \right) - 1}}{{x\left( {{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^5} + {{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^4} + ... + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{5184{x^5} + 5040{x^4} + 1900{x^3} + 345{x^2} + 30x}}{{x\left( {{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^5} + {{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^4} + ... + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{5184{x^4} + 5040{x^3} + 1900{x^2} + 345x + 30}}{{{{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^5} + {{\left( {\sqrt {1 + 4x} \sqrt[3]{{1 + 9x}}} \right)}^4} + ... + 1}}\\
= \dfrac{{30}}{{{1^5} + {1^4} + {1^3} + {1^2} + {1^1} + 1}}\\
= 5
\end{array}$
