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

lim_(x → π/4) (∫₂^(sec² x) f(t) dt)/(x² - π²/16) equals $\lim_{x \to \pi/4} \dfrac{\int_{2}^{\sec^2 x} f(t)\,dt}{x^2 - \dfrac{\pi^2}{16}}$ $\text{To find: the value of this limit}$ $\text{At } x = \dfrac{\pi}{4}:\ \sec^2\frac{\pi}{4} = 2 \Rightarrow \int_2^{2} f(t)\,dt = 0$ $x^2 - \dfrac{\pi^2}{16} = 0 \Rightarrow \dfrac{0}{0} \text{ form}$ $\text{L'Hôpital}: \lim_{x\to\pi/4} \dfrac{\frac{d}{dx}\int_2^{\sec^2 x} f(t)\,dt}{\frac{d}{dx}\left(x^2 - \frac{\pi^2}{16}\right)}$ $\frac{d}{dx}\int_2^{\sec^2 x} f(t)\,dt = f(\sec^2 x)\cdot\frac{d}{dx}(\sec^2 x)$ $\frac{d}{dx}(\sec^2 x) = 2\sec x\cdot\sec x\tan x = 2\sec^2 x\tan x$ $\frac{d}{dx}\left(x^2 - \frac{\pi^2}{16}\right) = 2x$ $\lim_{x\to\pi/4} \dfrac{f(\sec^2 x)\cdot 2\sec^2 x\tan x}{2x}$ $= \dfrac{f(2)\cdot 2\cdot 2\cdot 1}{2\cdot\frac{\pi}{4}} = \dfrac{4 f(2)}{\pi/2}$ $= 4 f(2)\cdot\dfrac{2}{\pi} = \dfrac{8}{\pi} f(2)$ $\boxed{\dfrac{8}{\pi} f(2)} \quad \text{option (1)}$