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(B) Given $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+	an t \sec t f(t)ight) d t$.Notice that the integrand is the derivative of the

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$3$

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(B) Given $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+	an t \sec t f(t)ight) d t$.Notice that the integrand is the derivative of the product $f(t) \sec t$,i.e.,$\frac{d}{dt}(f(t) \sec t) = f^{\prime}(t) \sec t + f(t) \sec t 	an t$.Thus,$g(x) = \int\limits_{x}^{\pi / 4} d(f(t) \sec t) = [f(t) \sec t]_{x}^{\pi / 4}$.$g(x) = f\left(\frac{\pi}{4}ight) \sec\left(\frac{\pi}{4}ight) - f(x) \sec x$.Substituting the given values $f\left(\frac{\pi}{4}ight) = \sqrt{2}$ and $\sec\left(\frac{\pi}{4}ight) = \sqrt{2}$,we get $g(x) = \sqrt{2} \cdot \sqrt{2} - f(x) \sec x = 2 - \frac{f(x)}{\cos x}$.Now,$\lim\limits_{x ightarrow(\pi/2)^{-}} g(x) = \lim\limits_{x ightarrow(\pi/2)^{-}} \left(2 - \frac{f(x)}{\cos x}ight)$.Since $f(\pi/2) = 0$ and $\cos(\pi/2) = 0$,this is a $0/0$ form.Applying $L$'Hopital's Rule: $\lim\limits_{x ightarrow(\pi/2)^{-}} \frac{f(x)}{\cos x} = \lim\limits_{x ightarrow(\pi/2)^{-}} \frac{f^{\prime}(x)}{-\sin x} = \frac{f^{\prime}(\pi/2)}{-\sin(\pi/2)} = \frac{1}{-1} = -1$.Therefore,$\lim\limits_{x ightarrow(\pi/2)^{-}} g(x) = 2 - (-1) = 3$.

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