Limit Evaluation of Function Defined Piecewise Using Squeeze Theorem

To evaluate the limit $\lim_{x \to 0} f(x)g(x)$, we will construct suitable bounds for $f(x)$ and $g(x)$ for $x \in (0,1)$ and then apply the Squeeze Theorem.

Step 1: Finding Bounds for $f(x)$

The function $f(x)$ is defined piecewise as $f(x) = \sqrt{n}$ for $x \in \left[ \frac{1}{n+1}, \frac{1}{n} \right)$, where $n \in \mathbb{N}$.

From the interval for $x$, we have: $\frac{1}{n+1} \le x < \frac{1}{n}$

Inverting the inequalities yields: $n < \frac{1}{x} \le n+1$

This implies: $\frac{1}{x} - 1 \le n < \frac{1}{x}$

Taking the square root throughout (since $n \ge 1$): $\sqrt{\frac{1}{x} - 1} \le \sqrt{n} < \sqrt{\frac{1}{x}}$

Since $f(x) = \sqrt{n}$, we establish the following inequality for all $x \in (0,1)$: $\sqrt{\frac{1}{x} - 1} \le f(x) < \sqrt{\frac{1}{x}}$

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Step 2: Finding Bounds for $g(x)$

We are given that $g(x)$ satisfies: $\int_{x^2}^{x} \sqrt{\frac{1-t}{t}} \, dt < g(x) < 2\sqrt{x}$

Let $I(x) = \int_{x^2}^{x} \sqrt{\frac{1-t}{t}} \, dt$. For any $t \in [x^2, x]$ where $x \in (0,1)$, we have: $x^2 \le t \le x < 1 \implies 1 - x \le 1 - t \le 1 - x^2$

Dividing by $\sqrt{t} > 0$ and taking square roots gives: $\sqrt{1-x} \cdot \frac{1}{\sqrt{t}} \le \sqrt{\frac{1-t}{t}} \le \sqrt{1-x^2} \cdot \frac{1}{\sqrt{t}}$

Integrating from $t = x^2$ to $t = x$: $\sqrt{1-x} \int_{x^2}^{x} \frac{1}{\sqrt{t}} \, dt \le I(x) \le \sqrt{1-x^2} \int_{x^2}^{x} \frac{1}{\sqrt{t}} \, dt$

Evaluating the integral $\int_{x^2}^{x} \frac{1}{\sqrt{t}} \, dt$: $\int_{x^2}^{x} t^{-1/2} \, dt = \left[ 2\sqrt{t} \right]_{x^2}^{x} = 2\sqrt{x} - 2x = 2\sqrt{x}(1 - \sqrt{x})$

Hence, the integral $I(x)$ is bounded below by: $I(x) \ge 2\sqrt{x}(1 - \sqrt{x})\sqrt{1-x}$

Combining this with the given upper bound for $g(x)$, we obtain: $2\sqrt{x}(1 - \sqrt{x})\sqrt{1-x} < g(x) < 2\sqrt{x}$

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Step 3: Finding Bounds for $f(x)g(x)$

Since both $f(x)$ and $g(x)$ are strictly positive for $x \in (0,1)$, we multiply their respective lower and upper bounds:

Lower Bound for $f(x)g(x)$: $f(x)g(x) > \sqrt{\frac{1}{x} - 1} \cdot 2\sqrt{x}(1 - \sqrt{x})\sqrt{1-x}$ $= \sqrt{\frac{1-x}{x}} \cdot 2\sqrt{x}(1 - \sqrt{x})\sqrt{1-x} = 2(1-x)(1-\sqrt{x})$

Upper Bound for $f(x)g(x)$: $f(x)g(x) < \sqrt{\frac{1}{x}} \cdot 2\sqrt{x} = 2$

Combining these bounds gives: $2(1-x)(1-\sqrt{x}) < f(x)g(x) < 2$

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Step 4: Applying the Squeeze Theorem

Taking the limit as $x \to 0^+$: $\lim_{x \to 0^+} 2(1-x)(1-\sqrt{x}) = 2(1-0)(1-0) = 2$ $\lim_{x \to 0^+} 2 = 2$

By the Squeeze Theorem: $\lim_{x \to 0} f(x)g(x) = 2$

Therefore, the correct option is C.