lim _{x rightarrow frac{pi}{4}} frac{int_2^{s | vedclass

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(A) The given limit is of the form $\frac{0}{0}$ as $x \rightarrow \frac{\pi}{4}$,since $\sec^2(\frac{\pi}{4}) = 2$. Applying $L$'$H$ôpital's rule,we

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$\frac{8}{\pi} f(2)$

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(A) The given limit is of the form $\frac{0}{0}$ as $x ightarrow \frac{\pi}{4}$,since $\sec^2(\frac{\pi}{4}) = 2$. Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$: Numerator: $\frac{d}{dx} \int_2^{\sec^2 x} f(t) dt = f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x) = f(\sec^2 x) \cdot 2 \sec x \cdot \sec x 	an x = 2 f(\sec^2 x) \sec^2 x 	an x$. Denominator: $\frac{d}{dx} (x^2 - \frac{\pi^2}{16}) = 2x$. Now,evaluate the limit as $x ightarrow \frac{\pi}{4}$: $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{2 f(\sec^2 x) \sec^2 x 	an x}{2x} = \frac{2 f(2) \cdot (\sqrt{2})^2 \cdot 1}{2(\frac{\pi}{4})} = \frac{2 f(2) \cdot 2}{\frac{\pi}{2}} = \frac{8 f(2)}{\pi}$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

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