Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \dfrac{{{n^2} + 2n}}{{3{n^2} + n + 1}} = \lim \dfrac{{\dfrac{{{n^2} + 2n}}{{{n^2}}}}}{{\dfrac{{3{n^2} + n + 1}}{{{n^2}}}}} = \lim \dfrac{{1 + \dfrac{2}{n}}}{{3 + \dfrac{1}{n} + \dfrac{1}{{{n^2}}}}} = \dfrac{{1 + 0}}{{3 + 0 + 0}} = \dfrac{1}{3}\\
b,\\
\mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 2}}{{2{x^2} + 5x + 2}} = \mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 2}}{{\left( {x + 2} \right)\left( {2x + 1} \right)}} = \mathop {\lim }\limits_{x \to - 2} \dfrac{1}{{2x + 1}} = \dfrac{1}{{2.\left( { - 2} \right) + 1}} = - \dfrac{1}{3}\\
c,\\
\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 8} - 3}}{{{x^2} + 2x - 3}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {x + 8} - 3} \right)\left( {\sqrt {x + 8} + 3} \right)}}{{\left( {{x^2} + 2x - 3} \right).\left( {\sqrt {x + 8} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 8} \right) - {3^2}}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {\sqrt {x + 8} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {\sqrt {x + 8} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {x + 3} \right)\left( {\sqrt {x + 8} + 3} \right)}}\\
= \dfrac{1}{{\left( {1 + 3} \right)\left( {\sqrt {1 + 8} + 3} \right)}} = \dfrac{1}{{24}}
\end{array}\)
