Evaluate $\lim_{x \to 1} \frac{\sqrt[3]{x} -1}{\sqrt{x} -1}$

Question

Evaluate $\lim_{x \to 1} \frac{\sqrt[3]{x} -1}{\sqrt{x} -1}$

I want to solve this limit by employing the strategy of introducing a new variable $t$ in such a way as to make the problem simpler.

I've tried using $t = \sqrt[3]{x} \Rightarrow \lim_{t \to 1} \frac{t -1}{\sqrt{t^3} -1}$ but I can't seem to manipulate the problem in to something simpler.

Can anybody give a hint?

Answer 1

We have$\require{cancel}$

$$\frac{\sqrt[3]x-1}{\sqrt x-1}=\frac{\cancel{(\sqrt[6]x-1)}(\sqrt[6]x+1)}{\cancel{(\sqrt[6] x-1)}(\sqrt[6]x^2+\sqrt[6]x+1)}\xrightarrow{x\to1}\frac23$$

Answer 2

We have: $\displaystyle \lim_{x \to 1} \dfrac{x^{1/3} - 1}{x^{1/2} - 1} = \dfrac{\displaystyle \lim_{x \to 1} \dfrac{x^{1/3} - 1}{x - 1}}{\displaystyle \lim_{x \to 1} \dfrac{x^{1/2} - 1}{x - 1}} = \dfrac{(x^{1/3})'(1)}{(x^{1/2})'(1)} = \dfrac{\frac{1}{3}}{\frac{1}{2}} = \dfrac{2}{3}$