Couldn't find cube root limit $\lim_{x\to1}{\frac{\sqrt[3]{x-1}}{\sqrt{x}-1}}$ with conjugate.

Question

I tried to multiply by the conjugate but couldn't get anything.

$$\lim_{x\to1}{\frac{\sqrt[3]{x-1}}{\sqrt{x}-1}}$$

Answer 1

HINT

Let $y^3=x-1\to 0$ then consider

$$\frac{y}{\sqrt {y^3+1}-1}= \frac{y (\sqrt {y^3+1}+1)}{y^3}$$

Answer 2

Hint

Let $\sqrt x=y,x=y^2$ to find

$$\dfrac{\sqrt[3]{y^2-1}}{y-1}=\sqrt[3]{y+1}\cdot\sqrt[3]{\dfrac{y-1}{(y-1)^3}}$$

Answer 3

Option :

$y=\sqrt x$; $y \rightarrow 1$;

$\dfrac{(y^2-1)^{1/3}}{y-1}=$

$\dfrac{(y-1)^{1/3}(y+1)^{1/3}}{(y-1)^{1/3}((y-1)^{1/3})^2}=$

$ \dfrac {(y+1)^{1/3}}{((y-1)^{1/3})^2}.$