Đáp án:
\(d) - \dfrac{1}{6}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 2 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}} \right) = + \infty \\
Do:\mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty \\
\mathop {\lim }\limits_{x \to - \infty } \left( { - 2 + \dfrac{3}{x} + \dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}} \right) = - 2\\
b)\mathop {\lim }\limits_{x \to + \infty } {x^4}\left( {5 - \dfrac{4}{{{x^2}}} + \dfrac{3}{{{x^4}}}} \right) = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } {x^4} = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {5 - \dfrac{4}{{{x^2}}} + \dfrac{3}{{{x^4}}}} \right) = 5\\
c)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3 - 4}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt {x + 3} + 2}} = \dfrac{1}{{\sqrt {1 + 3} + 2}} = \dfrac{1}{4}\\
d)\mathop {\lim }\limits_{x \to 7} \dfrac{{x + 2 - 9}}{{\left( {7 - x} \right)\left( {\sqrt {x + 2} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 7} \dfrac{{ - 1}}{{\sqrt {x + 2} + 3}} = \dfrac{{ - 1}}{{\sqrt {7 + 2} + 3}} = - \dfrac{1}{6}
\end{array}\)
