Limit Evaluation of Definite Integral Involving Differentiable Function

To evaluate the limit $\lim_{x \to 0} g(x)$, we start by simplifying the function $g(x)$ defined as:

$g(x) = \int_{x}^{\frac{\pi}{2}} \left[ f'(t) \csc t - \cot t \csc t f(t) \right] dt$

Notice that the integrand can be recognized as the derivative of the product $f(t) \csc t$ with respect to $t$:

$\frac{d}{dt}\left( f(t) \csc t \right) = f'(t) \csc t + f(t) (-\csc t \cot t) = f'(t) \csc t - \cot t \csc t f(t)$

Thus, the integral becomes:

$g(x) = \int_{x}^{\frac{\pi}{2}} \frac{d}{dt} \left( f(t) \csc t \right) dt$

Using the Fundamental Theorem of Calculus:

$g(x) = \left[ f(t) \csc t \right]_{x}^{\frac{\pi}{2}} = f\left(\frac{\pi}{2}\right) \csc\left(\frac{\pi}{2}\right) - f(x) \csc x$

Given that $f\left(\frac{\pi}{2}\right) = 3$ and $\csc\left(\frac{\pi}{2}\right) = 1$, we have:

$g(x) = 3 \cdot 1 - \frac{f(x)}{\sin x} = 3 - \frac{f(x)}{\sin x}$

Now, taking the limit as $x \to 0$:

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

The limit $\lim_{x \to 0} \frac{f(x)}{\sin x}$ can be rewritten by dividing the numerator and denominator by $x$:

$\lim_{x \to 0} \frac{f(x)}{\sin x} = \lim_{x \to 0} \frac{\frac{f(x) - f(0)}{x}}{\frac{\sin x}{x}}$

Since $f(0) = 0$ and $f$ is differentiable with $f'(0) = 1$, and $\lim_{x \to 0} \frac{\sin x}{x} = 1$:

$\lim_{x \to 0} \frac{f(x)}{\sin x} = \frac{f'(0)}{1} = 1$

Substituting this back into our expression for $\lim_{x \to 0} g(x)$:

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