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?

Given $\displaystyle \lim_{x\rightarrow 1}\frac{1}{2}\left[\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{2}}-1}\right] = \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\left(\frac{x^{\frac{1}{3}}-1}{x-1}\right)\times \left(\frac{x-1}{x^{\frac{1}{2}}-1}\right)\right]$
Now Using  The formula $\displaystyle \lim_{y\rightarrow 1}\frac{y^n-1}{y-1} = n$
So we get $\displaystyle \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\frac{x^{\frac{1}{3}}-1}{x-1}\right]\times \lim_{x\rightarrow 1}\left[\frac{x-1}{x^{\frac{1}{2}}-1}\right] =\frac{1}{2} \times \frac{1}{3}\times \frac{2}{1} = \frac{1}{3}$
