How to calculate the following limit: $\lim_{x\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)$?

Question

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 $\lim_{x\to+\infty} \sqrt[n]{x}=1$.

Answer 1

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 $0$.

Answer 2

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 $0$.

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$$n(\sqrt[n]x-1)=y\iff x=\left(1+\frac yn\right)^n.$$