giúp em với gấp gấp!!!!

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giúp em với gấp gấp!!!!
giup-em-voi-gap-gap

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Eirian 4 years 2021-04-26T11:36:18+00:00 2 Answers 9 views 0

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  1. Ben
    0
    2021-04-26T11:37:50+00:00
    Reply

    Đáp án:

    \(\lim\limits_{x\to -\infty}\sqrt[3]{x^{2021} + x^{2020} + 2020}=-\infty\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    \quad \lim\limits_{x\to -\infty}\sqrt[3]{x^{2021} + x^{2020} + 2020}\\
    = \lim\limits_{x\to -\infty}\left(\sqrt[3]{x^{2021}}\cdot \sqrt[3]{1 + \dfrac1x + \dfrac{2020}{x^{2021}}} \right)\\
    = \lim\limits_{x\to -\infty}\sqrt[3]{x^{2021}}\cdot \lim\limits_{x\to -\infty}\sqrt[3]{1 + \dfrac1x + \dfrac{2020}{x^{2021}}}\\
    = -\infty\cdot 1\\
    = -\infty
    \end{array}\)

  2. Vodka
    0
    2021-04-26T11:38:09+00:00
    Reply

        $\lim_{x \to -\infty}$ $\sqrt[3]{x^{2021}+x^{2020}+2020}$  

    = $\lim_{x \to -\infty}$ ( $\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 $\lim_{x \to -\infty}$ $\sqrt[3]{x^{2021}+x^{2020}+2020}$   = – ∞

     

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