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

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

• $\infty\times0$ does not give always 1 – zar Oct 23 '16 at 11:44
• 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
• – Workaholic Nov 1 '16 at 21:36
• – Martin Sleziak Jan 14 '17 at 17:32

## 4 Answers

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 $$\lim_{n \to \infty}\sqrt{n}(\sqrt[n]{x} - 1) = \lim_{n \to \infty}n(\sqrt[n]{x} - 1)\cdot\frac{1}{\sqrt{n}} = \log x \cdot 0 = 0$$

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 $0$.

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 $$\lim_{n\to\infty}\sqrt{n}(-1)=-\infty.$$

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

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

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