Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\lim \frac{{3{n^3} + 3n + 5}}{{{n^3} - 4n + 2}} = \lim \dfrac{{3 + \frac{3}{{{n^2}}} + \frac{5}{{{n^3}}}}}{{1 - \frac{4}{{{n^2}}} + \frac{2}{{{n^3}}}}} = \frac{{3 + 0 + 0}}{{1 - 0 + 0}} = 3\\
2,\\
\mathop {\lim }\limits_{x \to 2} \left( {{x^4} - 4{x^3} + 3x + 1} \right) = {2^4} - {4.2^3} + 3.2 + 1 = - 9\\
3,\\
a,\\
\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} + 3x + 1}}{{3{x^2} + 3x + 2}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{1 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{3 + \frac{3}{x} + \frac{2}{{{x^2}}}}} = \frac{{1 + 0 + 0}}{{3 + 0 + 0}} = \frac{1}{3}\\
b,\\
\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 9}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x + 3} \right) = 3 + 3 = 6\\
4,\\
a,\\
\mathop {\lim }\limits_{x \to \infty } \frac{{4{x^2} + x + 1}}{{2{x^2} + 5x + 2}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{4 + \frac{1}{x} + \frac{1}{{{x^2}}}}}{{2 + \frac{5}{x} + \frac{2}{{{x^2}}}}} = \frac{{4 + 0 + 0}}{{2 + 0 + 0}} = 2\\
b,\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \left( {x - 3} \right) = 2 - 3 = - 1
\end{array}\)
