Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + 2020} \right)\sqrt[3]{{1 - 3x}} - 2020}}{x} = - 2020\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + 2020} \right)\sqrt[3]{{1 - 3x}} - 2020}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}.\sqrt[3]{{1 - 3x}} + 2020\sqrt[3]{{1 - 3x}} - 2020}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}.\sqrt[3]{{1 - 3x}} + 2020.\left( {\sqrt[3]{{1 - 3x}} - 1} \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {x.\sqrt[3]{{1 - 3x}} + \frac{{2020.\left[ {\left( {1 - 3x} \right) - {1^3}} \right]}}{{x.\left( {{{\sqrt[3]{{1 - 3x}}}^2} + \sqrt[3]{{1 - 3x}}.1 + {1^2}} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left( {x.\sqrt[3]{{1 - 3x}} + \frac{{2020.\left( { - 3x} \right)}}{{x.\left( {{{\sqrt[3]{{1 - 3x}}}^2} + \sqrt[3]{{1 - 3x}} + 1} \right)}}} \right)\\
= \mathop {\lim }\limits_{x \to 0} \left( {x.\sqrt[3]{{1 - 3x}} - \frac{{6060}}{{{{\sqrt[3]{{1 - 3x}}}^2} + \sqrt[3]{{1 - 3x}} + 1}}} \right)\\
= 0.\sqrt[3]{{1 - 3.0}} - \frac{{6060}}{{{{\sqrt[3]{{1 - 3.0}}}^2} + \sqrt[3]{{1 - 3.0}} + 1}}\\
= 0 - \frac{{6060}}{3}\\
= - 2020
\end{array}\)
