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(C) The given integral is $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$. Notice that the inte

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$2$

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(C) The given integral is $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$. Notice that the integrand is the derivative of the product $f(t) \operatorname{cosec} t$ with respect to $t$,since $\frac{d}{dt} [f(t) \operatorname{cosec} t] = f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t$. Thus,$g(x) = [f(t) \operatorname{cosec} t]_x^{\pi / 2} = f(\frac{\pi}{2}) \operatorname{cosec}(\frac{\pi}{2}) - f(x) \operatorname{cosec} x$. Substituting the values $f(\frac{\pi}{2})=3$ and $\operatorname{cosec}(\frac{\pi}{2})=1$,we get $g(x) = 3 - \frac{f(x)}{\sin x}$. Now,$\lim _{x \rightarrow 0} g(x) = \lim _{x \rightarrow 0} (3 - \frac{f(x)}{\sin x}) = 3 - \lim _{x \rightarrow 0} \frac{f(x)}{\sin x}$. Using $L'H\hat{o}pital's$ rule,$\lim _{x \rightarrow 0} \frac{f(x)}{\sin x} = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{\cos x} = \frac{f^{\prime}(0)}{\cos 0} = \frac{1}{1} = 1$. Therefore,$\lim _{x \rightarrow 0} g(x) = 3 - 1 = 2$.

Let $f: R \rightarrow R$ be a differentiable function such that $f(0)=0$,$f(\frac{\pi}{2})=3$ and $f^{\prime}(0)=1$. If $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$ for $x \in (0, \frac{\pi}{2}]$,then $\lim _{x \rightarrow 0} g(x)=$

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