Key Concepts and Formulas

• L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Leibniz Integral Rule: For a function $F(x) = \int_{a(x)}^{b(x)} f(t)\,dt$, its derivative is $F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$.
• Properties of Definite Integrals: $\int_a^a f(t)\,dt = 0$.

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(t)\,dt} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$ We use $f(t)$ as the integrand to avoid confusion with the variable $x$ in the limit.

Step 1: Check for Indeterminate Form We need to evaluate the limits of the numerator and the denominator as $x \to \frac{\pi}{4}$.

• Denominator: As $x \to \frac{\pi}{4}$, the denominator approaches $\left(\frac{\pi}{4}\right)^2 - \frac{\pi^2}{16} = \frac{\pi^2}{16} - \frac{\pi^2}{16} = 0$.

• Numerator: As $x \to \frac{\pi}{4}$, the upper limit of the integral, $\sec^2 x$, approaches $\sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2$. Thus, the integral $\int_2^{\sec^2 x} f(t)\,dt$ approaches $\int_2^2 f(t)\,dt$. Since the upper and lower limits are the same, this integral evaluates to $0$. Therefore, the numerator $\frac{\pi}{4}\int_2^{\sec^2 x} f(t)\,dt$ approaches $\frac{\pi}{4} \cdot 0 = 0$.

Since both the numerator and the denominator tend to $0$, the limit is of the indeterminate form $\frac{0}{0}$, and L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule We need to find the derivatives of the numerator and the denominator with respect to $x$.

Let $g(x) = \frac{\pi}{4}\int_2^{\sec^2 x} f(t)\,dt$ be the numerator and $h(x) = x^2 - \frac{\pi^2}{16}$ be the denominator.

Step 3: Differentiate the Numerator using Leibniz Integral Rule To differentiate $g(x)$, we use the Leibniz Integral Rule. Let $G(x) = \int_2^{\sec^2 x} f(t)\,dt$. The upper limit of integration is $b(x) = \sec^2 x$, and the lower limit is $a(x) = 2$. The derivative of the upper limit is $b'(x) = \frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot (\sec x \tan x) = 2 \sec^2 x \tan x$. The derivative of the lower limit is $a'(x) = \frac{d}{dx}(2) = 0$.

By the Leibniz Integral Rule: $G'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$ $G'(x) = f(\sec^2 x) \cdot (2 \sec^2 x \tan x) - f(2) \cdot (0)$ $G'(x) = 2 f(\sec^2 x) \sec^2 x \tan x$ Now, differentiate the numerator $g(x) = \frac{\pi}{4} G(x)$: $g'(x) = \frac{\pi}{4} G'(x) = \frac{\pi}{4} \left( 2 f(\sec^2 x) \sec^2 x \tan x \right)$ $g'(x) = \frac{\pi}{2} f(\sec^2 x) \sec^2 x \tan x$

Step 4: Differentiate the Denominator The derivative of the denominator $h(x) = x^2 - \frac{\pi^2}{16}$ is: $h'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}\left(\frac{\pi^2}{16}\right) = 2x - 0 = 2x$

Step 5: Apply L'Hôpital's Rule to find the Limit Now we can rewrite the limit using the derivatives: $L = \mathop {\lim }\limits_{x \to {\pi \over 4}} \frac{g'(x)}{h'(x)} = \mathop {\lim }\limits_{x \to {\pi \over 4}} \frac{{\frac{\pi}{2} f(\sec^2 x) \sec^2 x \tan x}}{{2x}}$

Step 6: Evaluate the Limit Substitute $x = \frac{\pi}{4}$ into the expression:

• $\sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2$
• $\tan\left(\frac{\pi}{4}\right) = 1$
• $2x = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$

$L = \frac{{\frac{\pi}{2} f(2) \cdot 2 \cdot 1}}{{\frac{\pi}{2}}}$ $L = \frac{{\pi f(2)}}{{\frac{\pi}{2}}}$ $L = \pi f(2) \cdot \frac{2}{\pi}$ $L = 2f(2)$

Common Mistakes & Tips

• Confusing Integration Variable: Ensure the integration variable is distinct from the limit variable. Using $f(t)$ for the integrand is crucial.
• Incorrect Derivative of $\sec^2 x$: The chain rule must be applied correctly: $\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot (\sec x \tan x) = 2 \sec^2 x \tan x$.
• Forgetting the Constant Factor: The $\frac{\pi}{4}$ factor in the numerator must be carried through the differentiation process.

Summary

The problem involves evaluating a limit of an indeterminate form $\frac{0}{0}$. We applied L'Hôpital's Rule, which required differentiating the numerator and the denominator. The differentiation of the numerator, which contained a definite integral with a variable upper limit, was performed using the Leibniz Integral Rule. After finding the derivatives of both the numerator and the denominator, we substituted the limiting value of $x$ to find the value of the limit. The continuity of $f$ ensures that $f(2)$ is well-defined.

The final answer is $\boxed{2f(2)}$.