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I want to compute $\displaystyle \lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right)$

  • $\displaystyle \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{x}\right)$

\begin{align*} \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{x}\right)&= \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{\sqrt[3]{x}^{3}}\right)\\ &=\lim _{x\to 0^{+}}\left(\sqrt[3]{\frac{x}{x^3}}\right)\\ &=\lim _{x\to 0^{+}}\left(\sqrt[3]{\frac{1}{x^2}}\right)\\ &=\lim _{x\to 0^{+}}\left(\sqrt[3]{\frac{1}{(0^{+})^2}}\right)=+\infty\\ \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{x}\right)&=+\infty \end{align*}

  • $\displaystyle \lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right)$

$x\to 0^{-}\implies x<0 \implies (-x)>0\implies (-x)^{3}>0 \implies (-x)=\sqrt[3]{(-x)^3} $ \begin{align*} \lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right)&= \lim _{x\to 0^{-}}\left(\frac{-\sqrt[3]{x}}{-x}\right)\\ &=\lim _{x\to 0^{-}}\left(\frac{-\sqrt[3]{x}}{\sqrt[3]{(-x)^3}}\right)\\ \end{align*}

I'm stuck here; please correct me if am wrong Thanks in advance

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2 Answers 2

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$$ \frac{\sqrt[3]{x}}{x} = \frac{x^{\frac13}}{x} = \frac{1}{x^{\frac23}} $$ $$\lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim _{x\to 0}\left(\frac{1}{x^{\frac23}}\right) = +\infty$$

For completing your approach, let $y = -x$ $$ x < 0 \implies y > 0$$ $$\lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim_{y\to 0^{+}}\left(\frac{\sqrt[3]{-y}}{-y}\right) = \lim_{y\to 0^{+}}\left(\frac{-\sqrt[3]{y}}{-y}\right) = \lim_{y\to 0^{+}}\left(\frac{\sqrt[3]{y}}{y}\right) = +\infty $$ $$ \therefore \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right) =\lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right)= +\infty$$

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    $\begingroup$ Since $\sqrt[3]{x}$ then you supposed that x is positive am I wrong $\endgroup$ Nov 7, 2021 at 10:53
  • $\begingroup$ @Donfreecss You are confusing cube roots with square roots. For $\sqrt{x}$, $x$ has to be non-negative, but that isn't the case with $\sqrt[3]{x}$. Domain of $\sqrt[3]{x}$ is $\mathbb{R}$ $\endgroup$
    – Ankit Saha
    Nov 7, 2021 at 10:57
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    $\begingroup$ please, do you have any reference that talk about that name of book ? $\endgroup$ Nov 7, 2021 at 11:06
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    $\begingroup$ @Donfreecss Just think about it logically. The domain of $\sqrt{x}$ is $x \ge 0$ because the square of a real number is always non-negative. But the cube of a real number can be negative. For example, $(-2)^3 = -8$. The cube root of $-8$ is given by, $\sqrt[3]{-8} = -2$ $\endgroup$
    – Ankit Saha
    Nov 7, 2021 at 11:11
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First observe that we have an indetermination (0/0) for:

$$\displaystyle \lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right)$$

Then, we have just to apply L'hospital'Rule, which is simply to take the derivative of the numerator and denominator, this is:

$$\displaystyle \lim _{x\to 0}\left(\frac{\frac{x^{-2/3}}{3}}{x' = 1}\right) = \lim_{x \to 0} \left({\frac{x^{-2/3}}{3}}\right) = \infty$$

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    $\begingroup$ L'hospital is okay, but one doesn't even need this here, as one can just simplify $\endgroup$
    – LegNaiB
    Nov 7, 2021 at 10:52
  • $\begingroup$ I know, of course. $\endgroup$ Nov 7, 2021 at 10:53
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    $\begingroup$ You should write we could instead of we have ... $\endgroup$
    – PinkyWay
    Nov 7, 2021 at 10:59

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