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Question Number 203508 by Fridunatjan08 last updated on 20/Jan/24

$$Evaluatethegivenlimit:$$$$\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[n]{8}-1}{\sqrt[n]{16}-1}$$

Answered by esmaeil last updated on 20/Jan/24

$$=Y$$$$\frac{1}{n}=p\to (n\to \mathrm{\infty }\to p\to 0)$$$$\to Y=\underset{p\to 0}{\lim }\frac{2^{3p}-1}{2^{4p}-1}\stackrel{hopital}{=}\underset{p\to 0}{\lim }\frac{2^{3p}\times 3ln2}{2^{4p}\times 4ln2}$$$$=\frac{3}{4}$$

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