Let $f : (0,1) \to \mathbb{R}$ be the function defined as $f(x) = \sqrt{n}$ if $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$ where $n \in \mathbb{N}$. Let $g : (0,1) \to \mathbb{R}$ be a function such that $\int_{x^{2}}^{x} \sqrt{\frac{1-t}{t}}\,dt < g(x) < 2\sqrt{x}$ for all $x \in (0,1)$. Then $\lim_{x \to 0} f(x)g(x)$

Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [1/(n+1), 1/n) where n ∈ N. Let g : (0,1) → R be a function such that ∫(x²)ˣ √(1-t)/t dt < g(x) < 2√x for all x ∈ (0,1). Then lim(x → 0) f(x)g(x)

$f(x)=\sqrt{n},\quad x\in\left[\dfrac{1}{n+1},\dfrac{1}{n}\right)$

$\displaystyle\int_{x^2}^{x}\sqrt{\dfrac{1-t}{t}}\,dt < g(x) < 2\sqrt{x}$

$\lim_{x\to0} f(x)g(x) = ?$

$1.\ \text{relation between } n \text{ and } x$

$2.\ \text{upper bound: } 2\sqrt{nx}$

$3.\ \text{lower bound: } \sqrt{n}\,I(x)$

$\sqrt{n}\,I(x) < f(x)g(x) < 2\sqrt{nx},\quad I(x)=\int_{x^2}^{x}\sqrt{\dfrac{1-t}{t}}\,dt$

$x\in\left[\dfrac{1}{n+1},\dfrac{1}{n}\right) \Rightarrow \dfrac{1}{x}-1 \le n < \dfrac{1}{x}$

$\Rightarrow nx\in[1-x,\,1) \Rightarrow \sqrt{nx}\to 1$

$2\sqrt{nx} \to 2\cdot 1 = 2$

$\int\sqrt{\dfrac{1-t}{t}}\,dt = \sqrt{t(1-t)}+\sin^{-1}\sqrt{t}$

$I(x) = \left[\sqrt{t(1-t)}+\sin^{-1}\sqrt{t}\right]_{x^2}^{x}$

$I(x) = 2\sqrt{x} - x + o(\sqrt{x})$

$\sqrt{n}\,I(x) = 2\sqrt{nx} - x\sqrt{n} + o(1) \to 2$

$\sqrt{n}\,I(x) \to 2,\qquad 2\sqrt{nx}\to 2$

$\Rightarrow \lim_{x\to0} f(x)g(x) = 2$

$\lim_{x\to0}f(x)g(x) = 2$

$\boxed{2}\quad \text{option (3)}$