Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{4{x^3} + 2{x^2} - 2}}{{{x^3} + 5x - 1}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{{4{x^3} + 2{x^2} - 2}}{{{x^3}}}}}{{\dfrac{{{x^3} + 5x - 1}}{{{x^3}}}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{4 + \dfrac{2}{x} - \dfrac{2}{{{x^3}}}}}{{1 + \dfrac{5}{{{x^2}}} - \dfrac{1}{{{x^3}}}}} = \dfrac{{4 + 0 - 0}}{{1 + 0 - 0}} = 4\\
b,\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{4{x^2} + 7x - 30}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {4{x^2} - 8x} \right) + \left( {15x - 30} \right)}}{{x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{4x\left( {x - 2} \right) + 15\left( {x - 2} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {4x + 15} \right)}}{{x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \left( {4x + 15} \right) = 4.2 + 15 = 23
\end{array}\)
