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