For $x\in \left[\frac{1}{n+1},\frac{1}{n}\right)$ we have $\sqrt{n}\le \frac{1}{\sqrt{x}}<\sqrt{n+1}$, so as $x\to 0^{+}$, $f(x)\sim \frac{1}{\sqrt{x}}$. More precisely, $\sqrt{x}f(x)\to 1$.

Using $\int \sqrt{\frac{1-t}{t}}dt={\sin }^{-1}\sqrt{t}+\sqrt{t}\sqrt{1-t}+C$, the lower bound integral evaluates to

${\sin }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}-{\sin }^{-1}x-x\sqrt{1-x^2}$.

Divide the sandwich ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt{x}$ by $\sqrt{x}$ and take $x\to 0^{+}$:

The lower bound divided by $\sqrt{x}$ tends to $2$ (since ${\sin }^{-1}\sqrt{x}\sim \sqrt{x}$ and $\sqrt{x}\sqrt{1-x}\sim \sqrt{x}$), and the upper bound divided by $\sqrt{x}$ equals $2$.

Hence $\underset{x\to 0^{+}}{\lim }\frac{g(x)}{\sqrt{x}}=2$, and therefore $\underset{x\to 0^{+}}{\lim }f(x)g(x)=\underset{x\to 0^{+}}{\lim }\left(\sqrt{x}f(x)\right)\cdot \frac{g(x)}{\sqrt{x}}=1\cdot 2=2$.