I have to solve a series of limits and I can't find out this one. $$\lim_{n\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)= \,?$$ I have the feeling that this is equal to $0$, but I don't know how to prove it. Note that it may be a good idea to use $\lim_{x\to+\infty} \sqrt[n]{x}=1$.
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4Do you mean $n\rightarrow\infty$? – Wojowu Oct 18 '15 at 10:27
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Yes, I meant when n tends to infinity. sorry. – Brunner Nathan Oct 18 '15 at 12:30
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See also Limit $\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)$ – Martin Sleziak Jan 14 '17 at 17:29
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You may observe that, for any fixed $x>0$, assuming that $n \to \infty$, then $$ \sqrt[n]{x}=e^{1/n\times \log x}=1+O\left( \frac1n\right) $$ giving $$\lim_{n\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)=O\left( \frac1{\sqrt{n}}\right)$$ and the desired limit is equal to $0$.
Olivier Oloa
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@Wojowu I assume the OP means, for any fixed $x$, what is the limit as $n \to \infty$. Edited. Thanks. – Olivier Oloa Oct 18 '15 at 10:30
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If we assume this is what OP really means, then I agree that this is correct. – Wojowu Oct 18 '15 at 10:33
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Obviously for any $n>0$,
$$\lim_{x\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)=\infty,$$ so that we infer that the real question is
$$\lim_{n\to+\infty} \sqrt{n}(\sqrt[n]{x}-1).$$
Knowing that $$\lim_{n\to+\infty} n(\sqrt[n]{x}-1)=\ln(x),$$
the limit is $0$.
$$n(\sqrt[n]x-1)=y\iff x=\left(1+\frac yn\right)^n.$$