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

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

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*$\displaystyle \lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right)$
$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
 A: $$ \frac{\sqrt[3]{x}}{x} = \frac{x^{\frac13}}{x} = \frac{1}{x^{\frac23}} $$
$$\lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim _{x\to 0}\left(\frac{1}{x^{\frac23}}\right) = +\infty$$
For completing your approach, let $y = -x$
$$ x < 0 \implies y > 0$$
$$\lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim_{y\to 0^{+}}\left(\frac{\sqrt[3]{-y}}{-y}\right) = \lim_{y\to 0^{+}}\left(\frac{-\sqrt[3]{y}}{-y}\right) = \lim_{y\to 0^{+}}\left(\frac{\sqrt[3]{y}}{y}\right) = +\infty  $$
$$ \therefore \lim _{x\to 0^{+}}\left(\frac{\sqrt[3]{x}}{x}\right) = \lim _{x\to 0^{-}}\left(\frac{\sqrt[3]{x}}{x}\right) =\lim _{x\to 0}\left(\frac{\sqrt[3]{x}}{x}\right)= +\infty$$
A: 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:
$$\displaystyle \lim _{x\to 0}\left(\frac{\frac{x^{-2/3}}{3}}{x' = 1}\right) = \lim_{x \to 0} \left({\frac{x^{-2/3}}{3}}\right) = \infty$$
