Step-by-step Solution:

1. Analyze the integrand in g(x): The given function is: $g(x) = \int\limits_x^{\pi /2} [f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt$ The integrand is f'(t)cosec(t) - cot(t)cosec(t)f(t). Let's see if this is the derivative of a known function, particularly a product involving f(t). Consider the derivative of the product $f(t) \text{cosec}\,t$ with respect to t using the product rule (uv)' = u'v + uv'. Let u = f(t) and $v = \text{cosec}\,t$. Then u' = f'(t) and $v' = -\text{cosec}\,t \cot t$. ${d \over {dt}}(f(t)\text{cosec}\,t) = f'(t) \text{cosec}\,t + f(t)( - \text{cosec}\,t \cot t)$ ${d \over {dt}}(f(t)\text{cosec}\,t) = f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)$ This exactly matches the integrand.

2. Evaluate the integral: Since the integrand is the derivative of f(t) cosec(t), we can use the Fundamental Theorem of Calculus Part 2, which states \int_a^b F'(t) dt = F(b) - F(a)$. g(x) = \int\limits_x^{\pi /2} {d \over {dt}}(f(t)\text{cosec},t)dt$$g(x) = \left[ {f(t)\text{cosec},t} \right]_x^{\pi /2}$$g(x) = f\left( {{\pi \over 2}} \right)\text{cosec}\left( {{\pi \over 2}} \right) - f(x)\text{cosec},x$$

3. Substitute the given values: We are given f(π/2) = 3. We also know that cosec(π/2) = 1/sin(π/2) = 1/1 = 1. Substituting these values into the expression for g(x): $g(x) = (3)(1) - f(x) {1 \over {\sin x}}$ $g(x) = 3 - {{f(x)} \over {\sin x}}$

4. Calculate the limit: We need to find $\mathop {\lim }\limits_{x \to 0} g(x)$. ${\mathop {\lim }\limits_{x \to 0} g(x)} = \mathop {\lim }\limits_{x \to 0} \left( {3 - {{f(x)} \over {\sin x}}} \right)$ ${\mathop {\lim }\limits_{x \to 0} g(x)} = 3 - \mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}$ Let's evaluate the limit $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}$. We are given that f(0) = 0. Also, sin(0) = 0. Since f$ is differentiable, it is also continuous, so $\mathop {\lim }\limits_{x \to 0} f(x) = f(0) = 0$. This limit is of the indeterminate form $0/0.

5. Apply L'Hôpital's Rule: Since we have a 0/0 indeterminate form, we can apply L'Hôpital's Rule. ${\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}} = \mathop {\lim }\limits_{x \to 0} {{f'(x)} \over {\cos x}}$ Now, we can substitute x = 0. We are given f'(0) = 1 and we know cos(0) = 1. ${\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over {\cos x}}} = {{f'(0)} \over {\cos 0}} = {1 \over 1} = 1$ Alternatively, using the definition of the derivative: ${\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}} = \mathop {\lim }\limits_{x \to 0} {{f(x)/x} \over {\sin x / x}} = {{\mathop {\lim }\limits_{x \to 0} {{f(x) - f(0)} \over {x-0}}} \over {\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x}}} = {{f'(0)} \over 1} = {1 \over 1} = 1$

6. Find the final answer: Substitute the value of the limit back into the expression for $\mathop {\lim }\limits_{x \to 0} g(x)$. ${\mathop {\lim }\limits_{x \to 0} g(x)} = 3 - 1 = 2$ Therefore, the value of the limit is 2.