<s>[INST] Simplifying $\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1}$ 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. 
 [/INST] 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?
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