Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $f(0)=0$, $f\left(\frac{\pi}{2}\right)=3$ and $f'(0)=1$. If $g(x)=\int_{x}^{\frac{\pi}{2}}\left[f'(t)\operatorname{cosec}t-\cot t\operatorname{cosec}t\,f(t)\right]dt$ for $x\in\left(0,\frac{\pi}{2}\right]$, then $\lim_{x\to 0}g(x)=$ ______.

$f(0) = 0, \quad f\left(\dfrac{\pi}{2}\right) = 3, \quad f'(0) = 1$

$g(x) = \int_x^{\pi/2}\left[f'(t)\operatorname{cosec}t - \cot t\operatorname{cosec}t\,f(t)\right]dt$

$\text{To find: } \lim_{x \to 0} g(x)$

$\dfrac{d}{dt}\big(f(t)\operatorname{cosec}t\big) = f'(t)\operatorname{cosec}t - f(t)\cot t\operatorname{cosec}t$

$g(x) = \int_x^{\pi/2} \dfrac{d}{dt}\big(f(t)\operatorname{cosec}t\big)\,dt = \Big[f(t)\operatorname{cosec}t\Big]_x^{\pi/2}$

$g(x) = f\!\left(\tfrac{\pi}{2}\right)\operatorname{cosec}\tfrac{\pi}{2} - f(x)\operatorname{cosec}x = 3 - \dfrac{f(x)}{\sin x}$

$\lim_{x \to 0} g(x) = 3 - \lim_{x \to 0}\dfrac{f(x)}{\sin x}$

$\dfrac{f(x)}{\sin x} = \dfrac{f(x)}{x}\cdot\dfrac{x}{\sin x} \longrightarrow f'(0)\cdot 1 = 1$

$\lim_{x \to 0} g(x) = 3 - 1 = 2$

$\boxed{2}$