giúp em với gấp gấp!!!! Question giúp em với gấp gấp!!!! in progress 0 Môn Toán Eirian 4 years 2021-04-26T11:36:18+00:00 2021-04-26T11:36:18+00:00 2 Answers 9 views 0
Answers ( )
Đáp án:
\(\lim\limits_{x\to -\infty}\sqrt[3]{x^{2021} + x^{2020} + 2020}=-\infty\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to -\infty}\sqrt[3]{x^{2021} + x^{2020} + 2020}\\
= \lim\limits_{x\to -\infty}\left(\sqrt[3]{x^{2021}}\cdot \sqrt[3]{1 + \dfrac1x + \dfrac{2020}{x^{2021}}} \right)\\
= \lim\limits_{x\to -\infty}\sqrt[3]{x^{2021}}\cdot \lim\limits_{x\to -\infty}\sqrt[3]{1 + \dfrac1x + \dfrac{2020}{x^{2021}}}\\
= -\infty\cdot 1\\
= -\infty
\end{array}\)
$\lim_{x \to -\infty}$ $\sqrt[3]{x^{2021}+x^{2020}+2020}$
= $\lim_{x \to -\infty}$ ( $\sqrt[3]{x^{2021} }. \sqrt[3]{1+ 1/x +\frac{2020}{x^{2021}}} $ )
* $\lim_{n \to -\infty} ($ $\sqrt[3]{x^{2021}}$ )= -∞
* $\lim_{n \to -\infty}\sqrt[3]{1+1/x+\frac{2020}{x^{2021}}} $ = 1
vậy $\lim_{x \to -\infty}$ $\sqrt[3]{x^{2021}+x^{2020}+2020}$ = – ∞