I had this limit $$ \lim_{x \to \infty} \sqrt[n]{\frac{1}{n}-\frac{1}{2^n}} $$ and I got it to: $$ e^{\lim_{x \to \infty} \frac{1}{n}\ln{(\frac{1}{n}-\frac{1}{2^n})}} $$ Now looking at the limit I have $$\lim_{x \to \infty} \frac{1}{n}\ln{(\frac{1}{n}-\frac{1}{2^n})}$$ I don't see how to get rid of this $\ln$ or progress any further. Tried some substitutions but none seem to help. Managed to finish it with L'Hopital but formulation doesn't allow it.
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Hint: $\frac 1{2n} \leq \frac 1 n-\frac 1 {2^{n}} \leq \frac 1 n$, so $(\frac 1{2n})^{1/n} \leq (\frac 1 n-\frac 1 {2^{n}})^{1/n}\leq (\frac1 n)^{1/n}$ for all $n$. Use squeeze Theorem.
Proof of the fact that $n^{1/n} \to 1$ as $n \to \infty$ wihtout L'Hopital's Rule:
$n^{1/n} =e^{\frac {\ln n} n}$ and $\frac {\ln n} n \leq \frac {2\ln \sqrt n} n \leq \frac {2(\sqrt n -1)} n \to 0$ where I have used the inequality $\ln x \leq x-2$ valid for all $x \geq 1$.
Kavi Rama Murthy
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But how do i get limit of (1/n)^(1/n)? I'm encountering pretty similar problems like in the starting limit. – Comizard Sep 16 '21 at 12:38
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@Comizard Does this answer help? – user2661923 Sep 16 '21 at 13:23
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$n^{1/n}\to 1$ as $n\to \infty$. – Kavi Rama Murthy Sep 16 '21 at 13:28
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@Comizard I have edited my answer. Let me know if you want a proof of the inequality $\ln x \leq x-1$ for $x \geq 1$. – Kavi Rama Murthy Sep 16 '21 at 23:17