Đáp án:
4) \(\dfrac{1}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
3)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{4{x^2} + x + 5 - 4{x^2}}}{{\sqrt {4{x^2} + x + 5} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 5}}{{\sqrt {4{x^2} + x + 5} - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + \dfrac{5}{x}}}{{ - \sqrt {4 + \dfrac{1}{x} + \dfrac{5}{{{x^2}}}} - 2}} = - \dfrac{1}{{2 + 2}} = - \dfrac{1}{4}\\
4)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3 - 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {x + 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {1 + 1} \right)\left( {\sqrt {1 + 3} + 2} \right)}} = \dfrac{1}{8}
\end{array}\)
