Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - \sqrt[3]{{6x + 1}}}}{{{x^2}}} = 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - \sqrt[3]{{6x + 1}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {4x + 1} - \left( {2x + 1} \right)} \right) + \left( {\left( {2x + 1} \right) - \sqrt[3]{{6x + 1}}} \right)}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\frac{{4x + 1 - {{\left( {2x + 1} \right)}^2}}}{{\sqrt {4x + 1} + 2x + 1}} + \frac{{{{\left( {2x + 1} \right)}^3} - \left( {6x + 1} \right)}}{{{{\left( {2x + 1} \right)}^2} + \left( {2x + 1} \right)\sqrt[3]{{6x + 1}} + {{\sqrt[3]{{6x + 1}}}^2}}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\frac{{ - 4{x^2}}}{{\sqrt {4x + 1} + 2x + 1}} + \frac{{8{x^3} + 12{x^2}}}{{{{\left( {2x + 1} \right)}^2} + \left( {2x + 1} \right).\sqrt[3]{{6x + 1}} + {{\sqrt[3]{{6x + 1}}}^2}}}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{ - 4}}{{\sqrt {4x + 1} + 2x + 1}} + \frac{{8x + 12}}{{{{\left( {2x + 1} \right)}^2} + \left( {2x + 1} \right)\sqrt[3]{{6x + 1}} + {{\sqrt[3]{{6x + 1}}}^2}}}} \right]\\
= \frac{{ - 4}}{{\sqrt {4.0 + 1} + 2.0 + 1}} + \frac{{8.0 + 12}}{{{{\left( {2.0 + 1} \right)}^2} + \left( {2.0 + 1} \right).\sqrt[3]{{6.0 + 1}} + {{\sqrt[3]{{6.0 + 1}}}^2}}}\\
= \frac{{ - 4}}{2} + \frac{{12}}{3}\\
= 2
\end{array}\)
