Đáp án:
$\begin{array}{l}
A = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 1}}{{{x^2} + x + 4}} = \frac{{ - 2 + 1}}{{{{\left( { - 2} \right)}^2} - 2 + 4}} = - \frac{1}{6}\\
B = \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{{{\sin }^2}2x - 3\cos x}}{{\tan \,x}} = \frac{{{{\sin }^2}\left( {\frac{\pi }{3}} \right) - 3cos\frac{\pi }{6}}}{{\tan \frac{\pi }{6}}} = \frac{{ - 18 + 3\sqrt 3 }}{4}\\
C = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2{x^2} - x + 1} - \sqrt[3]{{2x + 3}}}}{{3{x^2} - 2}} = \frac{{\sqrt {2 - 1 + 1} - \sqrt[3]{{2 + 3}}}}{{3 - 2}} = \sqrt 2 - \sqrt[3]{5}\\
D = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1} - 2}}{{\sqrt[3]{{3x + 1}} - 2}} = \frac{{\sqrt 4 - 2}}{{\sqrt[3]{4} - 2}} = 0
\end{array}$
