Let f: (0, 1) → ℝ be the function defined as f(x) = n if x ∈ [...

Given,

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

And $g: \left(0, 1\right) \to \mathbb{R}$ be a function such that ${∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt < g\left(x\right) < 2\sqrt{x}$ for all $x \in \left(0, 1\right)$.

Now we need to solve $1$ sided limit here to get some answer, as $\lim_{x \to {0}^{-}}$ doesn’t exist here (not in domain)

As $\dfrac{1}{n + 1} \le x < \dfrac{1}{n}$

$⇒ n + 1 \ge \dfrac{1}{x} > n$

$⇒ n \ge \dfrac{1}{x} - 1 ⇒ \sqrt{n} \ge \sqrt{\dfrac{1}{x} - 1}$

So, let $f\left(x\right) = \sqrt{\left(\dfrac{1}{x}\right) - 1}$$where \left(\cdot\right) =$ least integer function

Now multiplying $f\left(x\right)$ in ${∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt < g\left(x\right) < 2\sqrt{x}$ and taking limit we get,

$\lim_{x \to {0}^{+}}{∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt \cdot \sqrt{\left(\dfrac{1}{x}\right) - 1} \le \lim_{x \to {0}^{+}}f\left(x\right) \cdot g\left(x\right) \le \lim_{x \to {0}^{+}}\sqrt{\left(\dfrac{1}{x}\right) - 1} \times 2\sqrt{x}$

Now $\lim_{x \to {0}^{+}}\sqrt{\left(\dfrac{1}{x}\right) - 1} \times 2\sqrt{x} = \lim_{x \to {0}^{+}}2\sqrt{x}\sqrt{\left[\dfrac{1}{x}\right]}\left(\dfrac{1}{x} \notin Z\right)$

$= \lim_{x \to {0}^{+}}2\sqrt{x\left(\dfrac{1}{x} - \left\{\dfrac{1}{x}\right\}\right)} = 2$

$= \lim_{x \to {0}^{+}}2\sqrt{1 - x\left(\dfrac{1}{x}\right)} = 2; \left(\dfrac{1}{x} \notin Z\right)$

So, $\lim_{x \to {0}^{+}}{∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt \cdot \sqrt{\dfrac{1}{x} - \left\{\dfrac{1}{x}\right\}} = \dfrac{{∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt \cdot \sqrt{1 - x\left\{\dfrac{1}{x}\right\}}}{\sqrt{x}}$

And $\lim_{x \to {0}^{+}}\dfrac{{∫}_{{x}^{2}}^{x}\sqrt{\dfrac{1 - t}{t}}dt}{\sqrt{x}} = \lim_{x \to {0}^{+}}\dfrac{\sqrt{\dfrac{1 - x}{x}} - 2x\sqrt{\dfrac{1 - {x}^{2}}{{x}^{2}}}}{\dfrac{1}{2\sqrt{x}}}$

{Using L-hospital rule}

$= \lim_{x \to {0}^{+}}2\sqrt{1 - x} - 4\sqrt{x} \cdot \sqrt{1 - {x}^{2}} = 2$

Similarly for $\dfrac{1}{x} \in Z$ is equal to $2$.