Đáp án:
1) 2
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - 5x - 4}}{{4x - 2}} = \dfrac{{{0^2} - 5.0 - 4}}{{4.0 - 2}} = 2\\
2)\mathop {\lim }\limits_{x \to - 1} \dfrac{{ - 6x + 3}}{{\sqrt {2x + 11} - 3x}}\\
= \dfrac{{ - 6\left( { - 1} \right) + 3}}{{\sqrt {2\left( { - 1} \right) + 11} - 3\left( { - 1} \right)}} = \dfrac{3}{2}\\
3)\mathop {\lim }\limits_{x \to 4} \dfrac{{x\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{x}{{x + 1}} = \dfrac{4}{{4 + 1}} = \dfrac{4}{5}\\
4)\mathop {\lim }\limits_{x \to 2} \dfrac{{4{x^2} - 5x - 6}}{{x\left( {2 - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {4x + 3} \right)}}{{x\left( {2 - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{ - 4x - 3}}{x} = \dfrac{{ - 4.2 - 3}}{2} = - \dfrac{{11}}{2}\\
5)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{9{x^2} + x + 1 - 9{x^2}}}{{\sqrt {9{x^2} + x + 1} - 3x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + \dfrac{1}{x}}}{{ - \sqrt {9 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} - 3}}\\
= \dfrac{1}{{ - 3 - 3}} = - \dfrac{1}{6}
\end{array}\)
