Limit $\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})$

Let $x=\frac 1 n$ and $f(x)=3^x-2^x$ so $$\lim_{n\to\infty}\sqrt{n}(\sqrt[n]{3}-\sqrt[n]{2})=\lim_{x\to0} \sqrt x\frac{3^x-2^x}{ x}=\lim_{x\to0} \sqrt x\frac{f(x)-f(0)}{ x}=\lim_{x\to0}\sqrt{x} f'(0)=0$$

$$\sqrt n(\sqrt[n] 3-\sqrt[n] 2)=\frac{\sqrt n(\sqrt[n] 3-\sqrt[n] 2)\left(\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}\right)}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}=\frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}$$ 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 $$\begin{align}\lim_{n\to\infty} \frac{\sqrt n}{3n}\le&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}\le\lim_{n\to\infty} \frac{\sqrt n}{n}\\0\le&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}\le 0\\&\lim_{n\to\infty} \frac{\sqrt n}{\sum_{k=0}^{n-1} 3^{k/n}2^{1-(k+1)/n}}=0\\&\lim_{n\to\infty}\sqrt n(\sqrt[n]{3}-\sqrt[n] 2)=0\end{align}$$