Answer
$\lim\limits_{x \to 1}\frac{\sqrt[3] x -1}{\sqrt{x}-1} = \frac{2}{3}$
Work Step by Step
$\frac{\sqrt[3] x -1}{\sqrt{x}-1}$
We can think of the numerator as a difference of squares:
$\sqrt[3] x -1 = (x^{1/6}-1)(x^{1/6}+1)$
We can think of the denominator as a difference of cubes:
$\sqrt{x} -1 = (x^{1/6}-1)(x^{1/3}+x^{1/6}+1)$
We can evaluate the limit:
$\lim\limits_{x \to 1}\frac{\sqrt[3] x -1}{\sqrt{x}-1}$
$= \lim\limits_{x \to 1}\frac{(x^{1/6}-1)(x^{1/6}+1)}{(x^{1/6}-1)(x^{1/3}+x^{1/6}+1)}$
$= \lim\limits_{x \to 1}\frac{x^{1/6}+1}{x^{1/3}+x^{1/6}+1}$
$= \frac{1^{1/6}+1}{1^{1/3}+1^{1/6}+1}$
$= \frac{2}{3}$
