Question

Let \(f : \mathbb{R} \to \mathbb{R}\) be a continuous function. Then
\(\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\frac{\pi}{4}\int_{2}^{\sec^2 x} f(x)\,dx}{x^2 - \frac{\pi^2}{16}}\)
is equal to :

• A. \(f(2)\)
• B. \(2f(2)\)
• C. \(2f\left(\sqrt{2}\right)\)
• D. \(4f(2)\)

Hint:

At \(x=\pi/4\) the integral has equal limits, so the expression is of the form \(0/0\); apply L'Hopital's rule with the Leibniz rule.

Answer 1

The Correct Option is B

At \(x=\pi/4\), \(\sec^2 x = 2\), so \(\int_2^{2} f = 0\) and the denominator is also 0: the limit is of the form \(0/0\).
Apply L'Hopital's rule. Differentiate the numerator using the Leibniz rule: \(\frac{d}{dx}\left[\frac{\pi}{4}\int_2^{\sec^2 x} f(t)\,dt\right] = \frac{\pi}{4} f(\sec^2 x)\cdot 2\sec^2 x\tan x\).
Differentiate the denominator: \(2x\).
Limit \(= \frac{\frac{\pi}{4} f(2)\cdot 2\cdot 2\cdot 1}{2\cdot\frac{\pi}{4}} = \frac{\pi f(2)}{\pi/2} = 2f(2)\).
Answer: (2) \(2f(2)\)