Problem Statement:

We are given two functions:

1. $f:(0,1)\to \mathbb{R}$ defined as:

   $$f(x)=\sqrt{n}\text{if}x\in [\frac{1}{n+1},\frac{1}{n}),n\in \mathbb{N}\mathrm{.}$$

2. $g:(0,1)\to \mathbb{R}$ such that for all $x\in (0,1)$:

   $${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt{x}\mathrm{.}$$

We are to evaluate the limit:

$$\underset{x\to 0}{\lim }f(x)g(x)$$

and determine which of the given options (A)-(D) is correct.

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Step 1: Analyze $f(x)$.

The function $f(x)$ is piecewise constant:

• For $x\in [\frac{1}{n+1},\frac{1}{n})$, $f(x)=\sqrt{n}$.

• As $x\to 0^{+}$, $x$ falls into intervals with larger $n$.

Behavior as $x\to 0^{+}$:

• For $x\in [\frac{1}{n+1},\frac{1}{n})$, $x\approx \frac{1}{n}$.

• Thus, $f(x)=\sqrt{n}\approx \frac{1}{\sqrt{x}}$.

Limit of $f(x)$:

$$\underset{x\to 0^{+}}{\lim }f(x)=\underset{n\to \mathrm{\infty }}{\lim }\sqrt{n}=\mathrm{\infty }\mathrm{.}$$

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Step 2: Analyze $g(x)$.

The function $g(x)$ is bounded by:

$${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt{x}\mathrm{.}$$

Lower Bound:

$${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt={\int }_{x^2}^x(\frac{1}{\sqrt{t}}-\sqrt{t})dt\mathrm{.}$$

Compute the integral:

$$\int (\frac{1}{\sqrt{t}}-\sqrt{t})dt=2\sqrt{t}-\frac{2}{3}{t}^{3\mathrm{/}2}\mathrm{.}$$

Evaluate from $x^2$ to $x$:

$$[2\sqrt{x}-\frac{2}{3}{x}^{3\mathrm{/}2}]-[2x-\frac{2}{3}{x}^3]=2\sqrt{x}-\frac{2}{3}{x}^{3\mathrm{/}2}-2x+\frac{2}{3}{x}^3\mathrm{.}$$

For $x\to 0^{+}$, the dominant term is $2\sqrt{x}$, so:

$${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt\approx 2\sqrt{x}\mathrm{.}$$

Upper Bound:

$$g(x)<2\sqrt{x}\mathrm{.}$$

Conclusion for $g(x)$:

$$g(x)\approx 2\sqrt{x}\text{as}x\to 0^{+}\mathrm{.}$$

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Step 3: Evaluate ${\lim }_{x\to 0}f(x)g(x)$.

From Step 1 and Step 2:

$$f(x)\approx \frac{1}{\sqrt{x}},g(x)\approx 2\sqrt{x}\mathrm{.}$$

Thus:

$$f(x)g(x)\approx \frac{1}{\sqrt{x}}\times 2\sqrt{x}=2.$$

Therefore:

$$\underset{x\to 0}{\lim }f(x)g(x)=2.$$

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Verification of Bounds:

To ensure the approximation is valid, consider $x\in [\frac{1}{n+1},\frac{1}{n})$:

• For $x\approx \frac{1}{n}$, $f(x)=\sqrt{n}$.

• The lower bound on $g(x)$ is approximately $2\sqrt{x}\approx \frac{2}{\sqrt{n}}$.

• Thus:

  $$f(x)g(x)\approx \sqrt{n}\times \frac{2}{\sqrt{n}}=2.$$

The upper bound $g(x)<2\sqrt{x}$ also leads to:

$$f(x)g(x)<\sqrt{n}\times \frac{2}{\sqrt{n}}=2.$$

Hence, the limit is squeezed to 2.

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Final Answer:

The limit is equal to 2. Therefore, the correct option is (C).

$$\boxed{{\displaystyle C}}$$