Let $f:(0,1)\to R$ be the function defined as $f\left(x\right)=\sqrt{n}$ if $x\in \frac{1}{n+1},\frac{1}{n}$, where $n\in N$.

Let $g:(0,1)\to R$ be a function such that

${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g\left(x\right)<2\sqrt{\mathrm{x }}$for all $x\in (0,1)$.

Then, which of the following statements is/are true?

The correct option is The limit is equal to 2.

Step 1: Understand the given function $f\left(x\right)$

The function $f:(0,1)\to R$ is defined piecewise as:

$f\left(x\right)=\sqrt{n}$ for $x\in (\frac{1}{n+1},\frac{1}{n}]$, where $n\in N$.

Step 2: Evaluate the integral bounds for $g\left(x\right)$

We are given the inequality for $g\left(x\right)$ as:

${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g\left(x\right)<2\sqrt{x}$ for all $x\in (0,1)$.

Let us analyze the integrand $I\left(x\right)={\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt$ as $x\to 0^{+}$.

For $t\in [x^2,x]$ where $x$ is small and positive, $1-t\approx 1$.

Thus, $\sqrt{\frac{1-t}{t}}\approx \frac{1}{\sqrt{t}}$.

Evaluating the main integral term:

${\int }_{x^2}^x\frac{1}{\sqrt{t}}dt={\left[2\sqrt{t}\right]}_{x^2}^x=2\sqrt{x}-2x$

Step 3: Apply the Squeeze Theorem to determine the behavior of $g\left(x\right)$

Dividing the inequality by $\sqrt{x}$ for $x>0$:

$\frac{2\sqrt{x}-2x}{\sqrt{x}}<\frac{g\left(x\right)}{\sqrt{x}}<2$

$2-2\sqrt{x}<\frac{g\left(x\right)}{\sqrt{x}}<2$

Taking the limit as $x\to 0^{+}$:

$\underset{x\to 0^{+}}{lim}(2-2\sqrt{x})=2$

By the Squeeze Theorem, we get:

$\underset{x\to 0^{+}}{lim}\frac{g\left(x\right)}{\sqrt{x}}=2$

Step 4: Evaluate the required limit involving $f\left(x\right)$ and $g\left(x\right)$

For $x\in (\frac{1}{n+1},\frac{1}{n}]$, we have $f\left(x\right)=\sqrt{n}$.

Since $1n+1<x\le \frac{1}{n}$, taking square roots and inverses gives:

$\sqrt{n}\le \frac{1}{\sqrt{x}}<\sqrt{n+1}$

Thus, $f\left(x\right)\approx \frac{1}{\sqrt{x}}$ as $x\to 0^{+}$.

Therefore, evaluating the limit product $\underset{x\to 0^{+}}{lim}f\left(x\right)g\left(x\right)$:

$\underset{x\to 0^{+}}{lim}f\left(x\right)g\left(x\right)=\underset{x\to 0^{+}}{lim}(\frac{1}{\sqrt{x}}·g(x\left)\right)=\underset{x\to 0^{+}}{lim}\frac{2\sqrt{x}}{\sqrt{x}}=2$

Hence, the limit is equal to 2.