<s>[INST] Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit

$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$

I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
 [/INST] Hint The radical expressions that occur, $\sqrt{x}$ and $\sqrt[3]{x}$, are both integer powers of $$u := \sqrt[6]{x},$$ and so we can write the expression in the limit as a rational expression in $u$:
$$\frac{\sqrt[3]{x} - 1}{2(\sqrt{x} - 1)} = \frac{(\sqrt[6]{x})^2 - 1}{2 ((\sqrt[6]{x})^3 - 1)} = \frac{u^2 - 1}{2(u^3 - 1)}.$$ (We've thus made a rationalizing substitution.) Can you simplify this?
</s>