Đáp án:
f) \(\dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 3} \right)\left( {x - 1} \right)}}{{\left( {2x + 1} \right)\left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3}}{{2x + 1}}\\
= \dfrac{{1 + 3}}{{2 + 1}} = \dfrac{4}{3}\\
c)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 + \dfrac{3}{{{x^2}}} - \dfrac{4}{{{x^3}}}}}{{ - 1 - \dfrac{1}{x} + \dfrac{1}{{{x^3}}}}} = - 2\\
e)\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}\left( {\dfrac{{1 - x - 1}}{{x + 1}}} \right)\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}.\dfrac{{ - x}}{{x + 1}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{x + 1}}\\
= \dfrac{{ - 1}}{{0 + 1}} = - 1\\
b)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 - x} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{x + 7 - 9}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 - x} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \left( { - \sqrt {x + 7} - 3} \right)\\
= - \sqrt {2 + 7} - 3 = - 6\\
d)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \sqrt {1 - \dfrac{1}{x}} + \sqrt {4 + \dfrac{1}{{{x^2}}}} }}{{2 + \dfrac{3}{x}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1 + 2}}{2} = \dfrac{1}{2}\\
f)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{4{x^2} - x - 4{x^2}}}{{\sqrt {4{x^2} - x} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x}}{{\sqrt {4{x^2} - x} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1}}{{ - \sqrt {4 - \dfrac{1}{x}} - 2}} = \dfrac{1}{4}
\end{array}\)
