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

Question

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

Answer 1

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

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

Answer 2

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:

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