Simplifying $\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1}$

Question

I was attempting to go through a worked solution from my lecturer but I can't really understand this step:

$\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1} = \lim_{n\to \infty} \frac{1}{\sqrt[n]{4}+\sqrt[n]{2}+1}$

What was actually even done here? And why?

Any help is appreciated.

Answer 1

Hint. Note that $$x^3-1=(x-1)(x^2+x+1)$$ which implies $$\sqrt[n]{2^3}-1=(\sqrt[n]{2}-1)(\sqrt[n]{2^2}+\sqrt[n]{2}+1).$$

Answer 2

One applies the factorisation $$x^3-1=(x-1)(x^2+x+1)$$ to $x=\sqrt[n]2$.

Answer 3

Hint: $$\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}=\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}\cdot\frac{\sqrt[n]{4} + \sqrt[n]{2} + 1}{\sqrt[n]{4} + \sqrt[n]{2} + 1}=\frac{\sqrt[n]{8} - 1}{\bigg(\sqrt[n]{8} - 1\bigg)\bigg(\sqrt[n]{4} + \sqrt[n]{2} + 1\bigg)}$$

Can you take it from here?

Answer 4

$\frac{2^{1/n}-1}{2^{3/n}-1} = \frac{2^{1/n}-1}{(2^{1/n}-1)*(2^{2/n}+2^{1/n}+1)}$

$= \lim_{n \to \infty}\frac{1}{(2^{2/n}+2^{1/n}+1)}$

= $\frac{1}{(1+1+1)} = 1/3$

Answer 5

Let $\sqrt[n]2-1=h\implies \sqrt[n]8=(\sqrt[n]2)^3=(h+1)^3$

Now $h\ne0$ as$h\to0$