Step-by-step Solution:

The problem asks to evaluate the limit: $L = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{\int\limits_2^{{{\sec }^2}x} {f(t)\,dt} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$

Step 1: Identify the form of the limit

First, we evaluate the numerator and the denominator as $x \to {\pi \over 4}$.

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

• Numerator: $N(x) = \int\limits_2^{{{\sec }^2}x} {f(t)\,dt}$. As $x \to {\pi \over 4}$, the upper limit of the integral becomes ${\sec ^2}\left( {{\pi \over 4}} \right) = {\left( {\sqrt 2 } \right)^2} = 2$. So, the integral becomes $\int\limits_2^2 {f(t)\,dt} = 0$.

Since both the numerator and the denominator approach 0, the limit is of the indeterminate form ${0 \over 0}$.

Step 2: Apply L'Hôpital's Rule

Since the limit is of the form ${0 \over 0}$, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to a} \frac{g(x)}{h(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.

$L = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ {d \over dx} \left( \int\limits_2^{{{\sec }^2}x} {f(t)\,dt} \right) } \over { {d \over dx} \left( {{x^2} - {{{\pi ^2}} \over {16}}} \right) }}$

Step 3: Differentiate the denominator

The derivative of the denominator is: ${d \over dx} \left( {{x^2} - {{{\pi ^2}} \over {16}}} \right) = 2x$

Step 4: Differentiate the numerator

To differentiate the numerator, we use the Leibniz integral rule (a consequence of the Fundamental Theorem of Calculus and the Chain Rule). If $G(x) = \int\limits_{a}^{u(x)} {f(t)\,dt}$, then $G'(x) = f(u(x)) \cdot u'(x)$.

In our case, $u(x) = \sec^2 x$. The derivative of $u(x)$ is: $u'(x) = {d \over dx} (\sec^2 x) = 2 \sec x \cdot {d \over dx}(\sec x) = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x$ So, the derivative of the numerator is: ${d \over dx} \left( \int\limits_2^{{{\sec }^2}x} {f(t)\,dt} \right) = f(\sec^2 x) \cdot (2 \sec^2 x \tan x)$

Step 5: Evaluate the limit

Now, we substitute the derivatives back into the limit expression: $L = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{f(\sec^2 x) \cdot (2 \sec^2 x \tan x)} \over {2x}}$ Now, we can substitute $x = {\pi \over 4}$ into the expression:

• $\sec^2({\pi \over 4}) = (\sqrt{2})^2 = 2$
• $\tan({\pi \over 4}) = 1$

$L = {{f(2) \cdot (2 \cdot 2 \cdot 1)} \over {2 \cdot {\pi \over 4}}}$ $L = {{4f(2)} \over {{\pi \over 2}}}$ $L = 4f(2) \cdot {2 \over \pi}$ $L = {8 \over \pi}f(2)$

Comparing this result with the given options, we find that it matches option A.

Final Answer: The value of the limit is ${8 \over \pi }f(2)$.