$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$

I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as $x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?

  • 6
    $\begingroup$ Note that $\sqrt[3]{x^3+x}= x\sqrt[3]{1 + \frac{1}{x^2}}$. Then cancel the $x$ here with the $x$ in the denominator and go from there. $\endgroup$ – Minus One-Twelfth May 17 at 20:00

$$\lim_{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}=\lim_{x\to-\infty} \sqrt[3]{\frac{x^3+x}{x^3}} =\lim_{x\to-\infty} \sqrt[3]{1+\frac{1}{x^2}} = \sqrt[3]{1+\lim_{x\to-\infty}\frac{1}{x^2}}=1$$

  • $\begingroup$ So check your problem again :) $\endgroup$ – Nosrati May 17 at 20:29
  • $\begingroup$ Thank you! I was looking at the wrong answer, I got 1 too at some point and I thought that I should get +infinity from the answer I was looking. I need to be more careful! :D $\endgroup$ – Novac May 17 at 20:31
  • $\begingroup$ You are welcome. $\endgroup$ – Nosrati May 17 at 20:32

\begin{align} \lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3+x}}{x} =& \lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3\cdot(1+1/x^2)}}{x} \\ =& \lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3}\cdot\sqrt[\,3]{(1+1/x^2)}}{x} \\ =& \lim_{x\to -\infty} \frac{x\cdot\sqrt[\,3]{(1+1/x^2)}}{x} \\ =& \lim_{x\to -\infty} {\sqrt[\,3]{(1+1/x^2)}} \\ =& {\sqrt[\,3]{\lim_{x\to -\infty} (1+1/x^2)}} \\ =& {\sqrt[\,3]{(1+\lim_{x\to -\infty} (1/x^2))}} \\ =& {\sqrt[\,3]{(1+0)}} \\ =& {\sqrt[\,3]{1}} \\ =& 1 \end{align}


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