Limit $\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})$ I've stumbled across this problem:$$\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})$$ Intuition tells me that the square root infinity is weaker than the nth root zero, also $\lim_{n\to\infty}\sqrt[n]{3n^{n/2}}=\lim_{n\to\infty}\sqrt[n]{2n^{n/2}}$ probably. Any way to prove it properly? I don't know Mr. L'Hospital yet, don't use him please
 A: Let $x=\frac 1 n$ and $f(x)=3^x-2^x$ so
$$\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})=\lim_{x\to0} \sqrt x\frac{3^x-2^x}{ x}=\lim_{x\to0} \sqrt x\frac{f(x)-f(0)}{ x}=\lim_{x\to0}\sqrt{x} f'(0)=0$$
A: $$\sqrt n(\sqrt[n] 3-\sqrt[n] 2)=\frac{\sqrt n(\sqrt[n] 3-\sqrt[n] 2)\left(\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}\right)}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}=\frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}$$
Note that $1\le3^{k/n}2^{1-(k+1)/n}\le 3$ for $0\le k\le n-1$, so the sum $S$ satisfies $n\le S\le3n $.
Therefore $$\begin{align}\lim_{n\to\infty} \frac{\sqrt n}{3n}\le&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}\le\lim_{n\to\infty} \frac{\sqrt n}{n}\\0\le&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}\le 0\\&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}=0\\&\lim_{n\to\infty}\sqrt n(\sqrt[n]{3}-\sqrt[n] 2)=0\end{align}$$
A: Elementary:
It is known that $$\lim_{n\to\infty}\frac{\sqrt[n]{a}-1}{\frac{1}{n}}=\ln a, a>0.$$
Follow$$\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})= \lim_{n\to\infty}\frac{\sqrt{n}}{n}(\frac{\sqrt[n]{3}-1}{\frac{1}{n}}-\frac{\sqrt[n]{2}-1}{\frac{1}{n}})=0(\ln3-\ln2)=0.$$
