Answer
$$\lim_{x\to1}\frac{x^{1/3}-1}{\sqrt x-1}=\frac{2}{3}$$
Work Step by Step
$$A=\lim_{x\to1}\frac{x^{1/3}-1}{\sqrt x-1}=\lim_{x\to1}\frac{(x^{{1/6}})^2-1^2}{(x^{1/6})^3-1^3}$$
$$A=\lim_{x\to1}\frac{(x^{1/6}-1)(x^{1/6}+1)}{(x^{1/6}-1)(x^{1/3}+x^{1/6}+1)}$$
$$A=\lim_{x\to1}\frac{x^{1/6}+1}{x^{1/3}+x^{1/6}+1}$$
$$A=\frac{1^{1/6}+1}{1^{1/3}+1^{1/6}+1}$$
$$A=\frac{1+1}{1+1+1}=\frac{2}{3}$$
