Đáp án:
$\begin{array}{l}
a)\lim 2\left( {\sqrt {{n^2} + 1} - n} \right)\\
= \lim \dfrac{{2.\left( {{n^2} + 1 - {n^2}} \right)}}{{\sqrt {{n^2} + 1} + n}}\\
= \lim \dfrac{2}{{\sqrt {{n^2} + 1} + n}}\\
= 0\\
b)\lim \dfrac{{{n^2} + 2n + 1}}{{{n^2}\sqrt n + 3}}\\
= \lim \dfrac{{\dfrac{1}{{\sqrt n }} + \dfrac{2}{{n\sqrt n }} + \dfrac{1}{{{n^2}\sqrt n }}}}{{1 + \dfrac{3}{{{n^2}\sqrt n }}}} = 0\\
c)\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1} - 2}}{{{x^2} - 9}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{x + 1 - 4}}{{\left( {\sqrt {x + 1} + 2} \right)\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{1}{{\left( {\sqrt {x + 1} + 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{1}{{\left( {\sqrt {3 + 1} + 2} \right)\left( {1 + 3} \right)}}\\
= \dfrac{1}{{16}}\\
d)\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {x + 7} - 3}}{{x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 7 - 9}}{{\left( {x - 2} \right)\left( {\sqrt {x + 7} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{\sqrt {x + 7} + 3}}\\
= \dfrac{1}{{\sqrt {2 + 7} + 3}} = \dfrac{1}{6}
\end{array}$
