Question

$ \displaystyle\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)

$\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int\limits_{2}^{\sec ^{2} x} f(t) d t}{x^{2}-\frac{\pi^{2}}{16}} [\frac{0}{0}$ form]
$=\displaystyle\lim _{x \rightarrow \pi / 4} \frac{f\left(\sec ^{2} x\right) 2 \sec x \sec x \tan x}{2 x}$
[using L' Hospital's rule]
$=\frac{2 f(2)}{\pi / 4}=\frac{8}{\pi} f(2)$