Let f: R rightarrow R be a continuous functio | vedclass

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(B) Let the limit be $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$.Since the

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$2 f(2)$

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(B) Let the limit be $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \cdot f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x)}{2x}$.Using the chain rule,$\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot \sec x 	an x = 2 \sec^2 x 	an x$.Substituting this back:$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \cdot f(\sec^2 x) \cdot 2 \sec^2 x 	an x}{2x}$.At $x = \frac{\pi}{4}$,$\sec^2 x = 2$,$	an x = 1$,and $x = \frac{\pi}{4}$.$L = \frac{\frac{\pi}{4} \cdot f(2) \cdot 2 \cdot 2 \cdot 1}{2 \cdot \frac{\pi}{4}} = \frac{\pi \cdot f(2)}{\frac{\pi}{2}} = 2 f(2)$.

Let $f: R \rightarrow R$ be a continuous function. Then $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$ is equal to :

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