Let $$f : (0, 1) \to \mathbb{R}$$ be the functions defined as $$f(x) = \sqrt{n}$$ if $$x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$$ where $$n \in \mathbb{N}$$. Let $$g : (0, 1) \to \mathbb{R}$$ be a function such that $$\int_{x^2}^{x} \sqrt{\frac{1-t}{t}} \, dt < g(x) < 2\sqrt{x}$$ for all $$x \in (0, 1)$$. Then $$\lim_{x \to 0} f(x)g(x)$$

The points $$1,\frac12,\frac13,\ldots$$ split the interval $$(0,1)$$ into the half-open intervals $$\left[\frac1{n+1},\frac1n\right)\;(n\in\mathbb N)$$. For $$x\in\left[\frac1{n+1},\frac1n\right)$$ we are told $$f(x)=\sqrt n.$$

---------------------------------------------------------------------
1. A relation between $$n$$ and $$x$$.
Because $$x\in\left[\frac1{n+1},\frac1n\right)$$ we have $$\frac1{n+1}\le x&lt\frac1n\; \Longrightarrow\; n\le\frac1x&lt n+1.$$ Multiplying by $$x$$ gives $$n\,x\le1&lt(n+1)\,x.$$ Hence $$1-\frac1{n+1}&lt n\,x\le1\qquad -(1)$$ and therefore $$\sqrt{1-\frac1{n+1}}&lt\sqrt{n\,x}\le1.\qquad -(2)$$

---------------------------------------------------------------------
2. Squeezing $$g(x)$$ as $$x\to0^+$$.
Define $$I(x)=\int_{x^{2}}^{x}\sqrt{\frac{1-t}{t}}\;dt,$$ so that $$I(x)&lt g(x)&lt2\sqrt x.$$ For $$0&lt t\le x&lt1$$ we have $$1-t\ge1-x,$$ hence $$\sqrt{\frac{1-t}{t}}\ge\sqrt{1-x}\,\frac1{\sqrt t}.$$ Therefore $$I(x)\ge\sqrt{1-x}\int_{x^{2}}^{x}\frac{dt}{\sqrt t} =\sqrt{1-x}\,[2\sqrt t]_{x^{2}}^{x} =2\sqrt{1-x}\,(\sqrt x-x).$$ Divide this and the upper bound by $$\sqrt x$$: $$2\sqrt{1-x}\,(1-\sqrt x)\;&lt\;\frac{g(x)}{\sqrt x}\;&lt\;2.$$ When $$x\to0^{+}$$, both the left and right bounds tend to $$2$$, so by the squeeze theorem $$\lim_{x\to0^{+}}\frac{g(x)}{\sqrt x}=2\;\;\Longrightarrow\;\; g(x)=2\sqrt x\,[1+\varepsilon(x)]$$ with $$\varepsilon(x)\to0$$ as $$x\to0^{+}.$$

---------------------------------------------------------------------
3. Limit of the product $$f(x)g(x)$$.
Using $$f(x)=\sqrt n$$ and the form of $$g(x)$$ just obtained, $$f(x)g(x)=\sqrt n\,g(x)=2\sqrt n\,\sqrt x\,[1+\varepsilon(x)] =2\sqrt{n\,x}\,[1+\varepsilon(x)].$$ From inequality $$(2)$$ we have $$\sqrt{1-\dfrac1{n+1}}\;\le\;\sqrt{n\,x}\;\le\;1.$$ Hence $$2\sqrt{1-\dfrac1{n+1}}\,[1+\varepsilon(x)] \;\le\;f(x)g(x) \;\le\;2[1+\varepsilon(x)].$$ Now let $$x\to0^{+}$$. Then $$n\to\infty$$ (because $$x\sim1/n$$) so the factor $$\sqrt{1-\dfrac1{n+1}}\to1,$$ and simultaneously $$\varepsilon(x)\to0.$$ Therefore both the lower and upper bounds tend to $$2$$, giving $$\boxed{\displaystyle\lim_{x\to0^{+}}f(x)g(x)=2}.$$

---------------------------------------------------------------------
Option C is the correct choice.