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?

Since :$$\left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) =x-1\\ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) =x-1$$
so we have
$$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt [ 3 ]{ x } -1 }{ 2\sqrt { x } -2 }  \right) = } \frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) \left( \sqrt { x } +1 \right)  }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right)  } = } \\ =\frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( x-1 \right) \left( \sqrt { x } +1 \right)  }{ \left( x-1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right)  }  } =\frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } =\frac { 1 }{ 3 }  $$
