Đáp án:
$\begin{array}{l}
a)\lim \frac{{\sqrt {4n} }}{{2n + 2020}} = \lim \frac{{2\sqrt n }}{{2n + 2020}} = \lim \frac{1}{{\sqrt n + 1010}} = 0\\
b)\lim \frac{{\sqrt {4{n^2} + n + 1} + n + 3}}{{3n + 2020}}\\
= \lim \frac{{\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} + 1 + \frac{3}{n}}}{{3 + \frac{{2020}}{n}}}\\
= \frac{{\sqrt 4 + 1}}{3} = 1\\
c)\mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left| {x - 3} \right|}}{{{x^2} - x - 6}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{x + 2}}\\
= \frac{1}{{3 + 2}} = \frac{1}{5}\\
d)\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{\left( {x - 2} \right)\left( {\sqrt {x + 7} - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {x + 7} - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{\sqrt {x + 7} - x}}\\
= \frac{{2 - 1}}{{\sqrt {2 + 7} - 2}} = 1
\end{array}$
