$\lim\limits_{x \rightarrow \pi / 4} \frac{\int\limits_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$, equals
Answer & explanation
Correct answer: option 1
Using L' Hospital's rule, we have
$\lim\limits_{x \rightarrow \pi / 4} \frac{\int\limits_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$
$=\lim\limits_{x \rightarrow \pi / 4} \frac{2 \sec ^2 x \tan x f\left(\sec ^2 x\right)}{2 x}=\frac{2 \times 2 \times 1 \times f(2)}{2 \times \frac{\pi}{4}}=\frac{8}{\pi} f(2)$