Limit $\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)$

Question

$$\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)$$ I have never previously encountered functional sequences, so I am a newbie in this matter. I know that intuitively, $x^{1/n}\to1$ and that $\infty\cdot0=1$. As a result, the limit should be 1. However, can this reasoning be considered correct and rigorous enough? I'm pretty sure that if I write this in my first year bachelor exam, my answer will be considered incorrect.

What is the more formal way of doing this? (I think that the squeeze theorem should work, but can't choose suitable bounds). Thanks in advance.

Hints: 1. For every $x>0$, $x^{1/n}=e^{(\log x)/n}$. 2. $e^t-1\sim t$ when $t\to0$. (Hence the limit of your sequence is $0$.) — Did, Oct 23 '16 at 11:45
You only need to know that for some fixed $x>0$ we have that $\lim \sqrt[n]{x}=1$. — Masacroso, Oct 23 '16 at 11:49
Related: $\displaystyle\lim_{n\to\infty} \sqrt{n}\left(\sqrt[n]{n}-1\right)$. — Workaholic, Nov 01 '16 at 21:36
See also: How to calculate the following limit: $\lim_{x\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)$? — Martin Sleziak, Jan 14 '17 at 17:32

score 2 · Answer 1 · answered Oct 27 '16 at 05:21

First of all note that the question makes sense only when $x \geq 0$. Also for $x = 0$ we can see that the limit diverges to $-\infty$. For $x > 0$ we use the standard result $$\lim_{n \to \infty}n(x^{1/n} - 1) = \log x\tag{1}$$ and then we have $$\lim_{n \to \infty}\sqrt{n}(\sqrt[n]{x} - 1) = \lim_{n \to \infty}n(\sqrt[n]{x} - 1)\cdot\frac{1}{\sqrt{n}} = \log x \cdot 0 = 0$$

score 1 · Answer 2 · answered Oct 23 '16 at 11:47

We have $$\lim_{n \to \infty} \sqrt{n} \left( x^{1/n} - 1\right) = \lim_{n \to \infty} \sqrt{n} \left( \exp\left( \frac{1}{n}\log(x)\right) - 1\right) = \lim_{n \to \infty} \sqrt{n} \left( \frac{1}{n}\log(x) + O\left( \frac{1}{n^2}\log^2(x)\right)\right)\\ = \lim_{n \to \infty} \left(\frac{1}{\sqrt{n}}\log(x) + O\left( \frac{1}{n^{3/2}}\log^2(x)\right)\right) = 0$$ using Taylor expansion of $\exp$ at $0$.

score 0 · Answer 3 · answered Oct 23 '16 at 11:54

Recalling the limit $\lim_{t\to0}\dfrac{e^t-1}{t}=1$, you can write (for every $x>0$)

$$ \begin{align*} \lim_{n\to\infty}\sqrt{n}(\sqrt[n]{x}-1)&= \lim_{n\to\infty}\sqrt{n}\left(x^\frac{1}{n}-1\right)\\ &=\lim_{n\to\infty}\sqrt{n}\left(e^\frac{\ln x}{n}-1\right)\\ &=\lim_{n\to\infty}\sqrt{n}\frac{e^\frac{\ln x}{n}-1}{\frac{\ln x}{n}}\cdot\frac{\ln x}{n}\\ &=\lim_{n\to\infty}\frac{e^\frac{\ln x}{n}-1}{\frac{\ln x}{n}}\cdot\frac{\ln x}{\sqrt{n}}\\ &=0. \end{align*}\\ $$

If $x=0$ you have $$ \lim_{n\to\infty}\sqrt{n}(-1)=-\infty. $$

score 0 · Answer 4 · answered Oct 23 '16 at 11:57

Taking $\;n\;$ as a continuous variable and using l'Hospital (and assuming $\;x>0\;$):

$$\lim_{n\to\infty}\sqrt n\left(\sqrt[n]x-1\right)=\lim_{n\to\infty}\frac{x^{1/n}-1}{n^{-1/2}}\stackrel{\text{l'H}}=\lim_{n\to\infty}\frac{\left(-\frac1{n^2}\right)x^{1/n}\log x}{-\frac12n^{-3/2}}=$$

$$=2\log x\lim_{n\to\infty}\frac{x^{1/n}}{\sqrt n}=0\;,\;\;\text{since}\;\;x^{1/n}\xrightarrow[n\to\infty]{}1$$

Limit $\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)$

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