Lim (√(4x+1)-∛(6x+1))/x^2 khi x tiến đến 0

Đáp án:

\[4\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1}  – \sqrt[3]{{6x + 1}}}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {4x + 1}  – \left( {2x + 1} \right)} \right) + \left[ {\left( {2x + 1} \right) – \sqrt[3]{{6x + 1}}} \right]}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{4x + 1 – 4{x^2} – 4x – 1}}{{\sqrt {4x + 1}  + 2x + 1}} + \frac{{8{x^3} + 12{x^2} + 6x + 1 – 6x – 1}}{{2x + 1 + \sqrt[3]{{6x + 1}}}}}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{ – 4}}{{\sqrt {4x + 1}  + 2x + 1}} + \frac{{8x + 12}}{{2x + 1 + \sqrt[3]{{6x + 1}}}}} \right]\\
 = \frac{{ – 4}}{{\sqrt 1  + 1}} + \frac{{12}}{{1 + \sqrt[3]{1}}} = \frac{{ – 4}}{2} + \frac{{12}}{2} = 4
\end{array}\)