Question

$lim_ { x \to \frac{\pi}{4}} \frac{ \int \limits_2^{sec^2 \, x} \, f \, (t) \, dt }{ x^2 - \frac{\pi^2}{ 16}}$ equals

• A. $\frac{8}{ \pi} f$
• B. $\frac{2}{ \pi} f$
• C. $\frac{2}{ \pi} f \bigg( \frac{1}{2}\bigg)$
• D. $4f (2)$

Answer 1

Answer: (A)

$\frac{8}{ \pi} f$

Answer 2

$lim_ { x \to \frac{\pi}{4}} \frac{ \int \limits_2^{sec^2 \, x} \, f \, (t) \, dt }{ x^2 - \frac{\pi^2}{ 16}}$
= $lim_ { x \to \pi / 4} \frac{ f (sec^2 \, x) 2 \, sec \, x \, sec \, x \, tan \, x}{ 2x}$ \hspace25mm $\bigg [ \frac{0}{0} form \bigg ]$
\hspace25mm [using L' Hospital's rule]
= $\frac{2 f (2)}{ \pi / 4} = \frac{8}{ \pi}$ f (2)