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?
 A: 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?
A: hint: Let $x = t^6$, and simplify to a nicer expression.
A: Why not to apply L-Hospital's rule for  $\frac{0}{0}$ form 
$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{2\sqrt x-2}$$
$$=\lim_{x\to 1}\frac{\frac{d}{dx}\left(\sqrt[3]{x}-1\right)}{\frac{d}{dx}(2\sqrt x-2)}$$
$$=\lim_{x\to 1}\frac{\frac{1}{3}x^{-2/3}}{2\frac{1}{2}x^{-1/2}}$$
$$=\lim_{x\to 1}\frac{1}{3}x^{-1/6}=\frac{1}{3}$$
A: 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 $\displaystyle \frac{1}{2}\times \lim_{x\rightarrow 1}\left[\frac{x^{\frac{1}{3}}-1}{x-1}\right]\times \lim_{x\rightarrow 1}\left[\frac{x-1}{x^{\frac{1}{2}}-1}\right] =\frac{1}{2} \times \frac{1}{3}\times \frac{2}{1} = \frac{1}{3}$
A: 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
$$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt [ 3 ]{ x } -1 }{ 2\sqrt { x } -2 }  \right) = } \frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) \left( \sqrt { x } +1 \right)  }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right)  } = } \\ =\frac { 1 }{ 2 } \lim _{ x\rightarrow 1 }{ \frac { \left( x-1 \right) \left( \sqrt { x } +1 \right)  }{ \left( x-1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right)  }  } =\frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } =\frac { 1 }{ 3 }  $$
A: Hint:
$$\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right) = \frac{1}{2}\lim _{x\to 1}\left(\left(\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\right)\left(\frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}\right)\left(\frac{\sqrt{x} + 1}{\sqrt{x} + 1}\right)\right)$$
Now use the following equation to simplify:
$$(\sqrt[3]{x}-1)(\sqrt[3]{x^2} + \sqrt[3]{x} + 1) = (\sqrt{x}-1)(\sqrt{x}+1) = x - 1.$$
