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Step-by-Step Solution
Step 1: Recognize the Indeterminate Form
As \(x \to \tfrac{\pi}{4}\), the upper limit of the integral \(\sec^2 x\) becomes \(\sec^2 \tfrac{\pi}{4} = 2\). Hence,
\[
\int_{2}^{\sec^2 x} f(x)\,dx \to \int_{2}^{2} f(x)\,dx = 0.
\]
Meanwhile,
\[
x^2 - \frac{\pi^2}{16} \to \left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = 0.
\]
Thus, the overall expression is of the form \(\frac{0}{0}\), indicating an indeterminate form suitable for L’Hôpital’s Rule.
Step 2: Apply L’Hôpital’s Rule
According to L’Hôpital’s Rule, if \(\lim_{x \to a} \frac{g(x)}{h(x)}\) is \(\frac{0}{0}\), then
\[
\lim_{x \to a} \frac{g(x)}{h(x)}
= \lim_{x \to a} \frac{g'(x)}{h'(x)},
\]
provided the limit on the right side exists. Here, set
\[
g(x) = \frac{\pi}{4} \int_{2}^{\sec^2 x} f(t)\,dt,
\quad
h(x) = x^2 - \frac{\pi^2}{16}.
\]
Step 3: Differentiate the Numerator \( g(x) \)
Using the Fundamental Theorem of Calculus and the chain rule:
\[
\frac{d}{dx} \left[\int_{2}^{\sec^2 x} f(t)\,dt\right] = f(\sec^2 x) \times \frac{d}{dx}(\sec^2 x).
\]
Next,
\[
\frac{d}{dx}(\sec^2 x) = 2\sec x \sec x\tan x = 2\,\sec^2 x \tan x.
\]
Therefore,
\[
g'(x) = \frac{d}{dx}\biggl(\tfrac{\pi}{4}\biggr) \int_{2}^{\sec^2 x} f(t)\,dt
= \frac{\pi}{4} \cdot f(\sec^2 x) \cdot 2\,\sec^2 x \tan x.
\]
Step 4: Differentiate the Denominator \( h(x) \)
\[
h(x) = x^2 - \frac{\pi^2}{16}
\quad\Longrightarrow\quad
h'(x) = 2x.
\]
Step 5: Form the New Limit and Evaluate at \( x = \frac{\pi}{4} \)
By L’Hôpital’s Rule,
\[
\lim_{x \to \frac{\pi}{4}} \frac{g(x)}{h(x)}
= \lim_{x \to \frac{\pi}{4}} \frac{g'(x)}{h'(x)}
= \lim_{x \to \frac{\pi}{4}} \frac{\frac{\pi}{4} \cdot f(\sec^2 x) \cdot 2\,\sec^2 x \tan x}{2x}.
\]
Simplify the fraction:
\[
= \lim_{x \to \frac{\pi}{4}} \frac{\pi}{4} \cdot f(\sec^2 x)\,\sec^2 x\,\tan x \,\frac{2}{2x}
= \lim_{x \to \frac{\pi}{4}} \frac{\pi}{4} \cdot f(\sec^2 x)\, \sec^2 x \tan x \,\frac{1}{x}.
\]
Now, plug in \( x = \frac{\pi}{4} \):
\[
\sec^2 \bigl(\tfrac{\pi}{4}\bigr) = 2,\quad
\tan \bigl(\tfrac{\pi}{4}\bigr) = 1,\quad
\frac{1}{x} \Bigm|_{x=\tfrac{\pi}{4}} = \frac{4}{\pi}.
\]
So,
\[
= \frac{\pi}{4} \cdot f(2) \cdot 2 \cdot 1 \cdot \frac{4}{\pi}
= f(2)\cdot 2 \cdot \frac{\pi}{4}\cdot \frac{4}{\pi}
= 2 f(2).
\]
Final Answer
The limit equals \(2 f(2)\).
