1. Understand the piecewise function $f(x)$ near $0$

If $x\in \left[\frac1{n+1},\frac1n\right),$ then by definition $f(x)=\sqrt n.$

As $x\to 0^+$, we have $n\to \infty$. Also, for such $x$, $\frac1{n+1}\le x<\frac1n.$ Taking reciprocals, $n<\frac1x\le n+1.$ Hence $nx<1\le (n+1)x.$ So for small $x$, $n\sim \frac1x$, and therefore $f(x)=\sqrt n \sim \frac1{\sqrt x}.$

More precisely, $\sqrt{nx}\le f(x)\sqrt x<\sqrt{(n+1)x}.$ Since $nx<1\le (n+1)x$ and both tend to $1$ as $x\to 0$, we get $\lim_{x\to 0} f(x)\sqrt x=1.$

2. Estimate $g(x)$

We are given $\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}\,dt<g(x)<2\sqrt x,\qquad x\in(0,1).$

To find the limit of $f(x)g(x)$, it is enough to first find the asymptotic behavior of $g(x)$.

Consider $I(x)=\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}\,dt.$ Rewrite the integrand as $\sqrt{\frac{1-t}{t}}=\frac{\sqrt{1-t}}{\sqrt t}.$ For $t\to 0^+$, $\sqrt{1-t}\to 1,$ so the integrand behaves like $\frac1{\sqrt t}.$ Thus we expect $I(x)\sim \int_{x^2}^{x}\frac{dt}{\sqrt t}=2(\sqrt x-x).$ Hence $I(x)\sim 2\sqrt x$.

Let us prove this rigorously by dividing by $2\sqrt x$.

Since $0<t<x<1$, we have $0<\sqrt{1-t}<1$, so $0<I(x)<\int_{x^2}^{x}\frac{dt}{\sqrt t}=2(\sqrt x-x)<2\sqrt x.$ Thus the lower bound is already very close to the upper bound.

Now compute

Because on $[x^2,x]$, $\sqrt{1-t}\to 1$ uniformly as $x\to 0$, we get $I(x)=\int_{x^2}^{x}\frac{\sqrt{1-t}}{\sqrt t}\,dt \sim \int_{x^2}^{x}\frac{dt}{\sqrt t}=2(\sqrt x-x).$ Therefore $\frac{I(x)}{2\sqrt x}\to 1.$

Since $I(x)<g(x)<2\sqrt x,$ dividing by $2\sqrt x>0$ gives $\frac{I(x)}{2\sqrt x}<\frac{g(x)}{2\sqrt x}<1.$ By squeeze theorem, $\lim_{x\to 0}\frac{g(x)}{2\sqrt x}=1.$ Hence $g(x)\sim 2\sqrt x.$ So, $\lim_{x\to 0}\frac{g(x)}{\sqrt x}=2.$

3. Now compute $\lim_{x\to 0} f(x)g(x)$

Write $f(x)g(x)=\big(f(x)\sqrt x\big)\cdot \frac{g(x)}{\sqrt x}.$ We found

and $\lim_{x\to 0} \frac{g(x)}{\sqrt x}=2.$ Therefore $\lim_{x\to 0} f(x)g(x)=1\cdot 2=2.$

4. Check options

• A: does NOT exist — false
• B: is equal to $1$ — false
• C: is equal to $2$ — true
• D: is equal to $3$ — false

So the correct option is $\boxed{\text{C}}.$