Question about limit! How can I find this limit, 

$$\lim_{x\rightarrow1}\frac{x^{1/3}-1}{x^{1/2}-1}.$$

I know how to find it using the L'Hospital theorem. I would like to find it without it.
Let $z:=x^{1/6}$. It also tends to $1$ as $x\to 1$. Then we have
$$\frac{z^2-1}{z^3-1}=\frac{z+1}{z^2+z+1}\to\frac23$$
as $z\to 1$.
Divide by x-1 in numerator and denominator then give limit saperately
lim x→1 (x^(1/3)) - 1)(x-1)/lim x→1 ((x^(1/2) - 1)/(x-1) = (1/3)/(1/2) = 2/3