1. Key Concepts and Formulas

• Leibniz's Rule for Differentiation Under the Integral Sign: If $G(x) = \int_{a(x)}^{b(x)} h(t, x) dt$, then $G'(x) = h(b(x), x) \cdot b'(x) - h(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} h(t, x) dt$. For $G(x) = \int_0^x h(t) dt$, this simplifies to $G'(x) = h(x)$.
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Taylor Series Expansion of $\cos x$ around $x=0$: $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$

2. Step-by-Step Solution

We are given the integral equation: $\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } \quad \text{for } 0 \le x \le 1$ with the conditions $f(0) = 0$, $f(x) \ge 0$ for $x \in [0, 1]$, and $f$ is twice differentiable in $(0, 1)$. We need to find $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}$.

Step 1: Differentiate both sides of the integral equation using Leibniz's Rule. To eliminate the integrals and obtain a differential equation, we differentiate both sides of the given equation with respect to $x$. Applying Leibniz's Rule (specifically, $\frac{d}{dx} \int_0^x h(t) dt = h(x)$), we get: $\sqrt {1 - {{(f'(x))}^2}} = f(x)$ This equation implies that $f(x) \ge 0$ (since the left side is a square root) and $1 - (f'(x))^2 \ge 0$, which means $(f'(x))^2 \le 1$, or $-1 \le f'(x) \le 1$.

Step 2: Square both sides and rearrange to form a separable differential equation. Squaring both sides of $\sqrt {1 - {{(f'(x))}^2}} = f(x)$ gives: $1 - {(f'(x))^2} = {f^2}(x)$ Rearranging the terms to isolate $(f'(x))^2$: ${(f'(x))^2} = 1 - {f^2}(x)$ Taking the square root of both sides yields: $f'(x) = \pm \sqrt{1 - {f^2}(x)}$ Since $f(x) \ge 0$, we consider the differential equation in the form $\frac{df}{dx} = \pm \sqrt{1 - f^2}$. We can separate the variables: $\frac{df}{\sqrt{1 - f^2}} = \pm dx$

Step 3: Integrate the differential equation. Integrating both sides: $\int \frac{df}{\sqrt{1 - f^2}} = \int \pm dx$ The integral on the left is a standard form: $\int \frac{du}{\sqrt{1-u^2}} = \arcsin(u) + C$. Thus, we have: $\arcsin(f(x)) = \pm x + C$

Step 4: Use the initial condition to find the constant of integration. We are given $f(0) = 0$. Substituting $x=0$ into the equation: $\arcsin(f(0)) = \pm 0 + C$ $\arcsin(0) = C$ Since $\arcsin(0) = 0$, we get $C=0$. So, the equation becomes: $\arcsin(f(x)) = \pm x$ This implies $f(x) = \sin(\pm x)$. Since $\sin(-x) = -\sin(x)$, we have $f(x) = \pm \sin(x)$. Given that $f(x) \ge 0$ for $x \in [0, 1]$, and $\sin(x) \ge 0$ for $x \in [0, 1]$ (as $1$ radian is in the first quadrant), we must choose the positive sign. Therefore, $f(x) = \sin(x)$.

Step 5: Evaluate the limit. We need to find $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}$. Substituting $f(t) = \sin t$: $L = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {\sin t\,dt}$ First, evaluate the definite integral: $\int_0^x {\sin t\,dt} = [-\cos t]_0^x = -\cos x - (-\cos 0) = -\cos x + 1 = 1 - \cos x$ Now, substitute this back into the limit: $L = \mathop {\lim }\limits_{x \to 0} {{1 - \cos x} \over {{x^2}}}$ As $x \to 0$, the expression is of the indeterminate form $\frac{0}{0}$. We can apply L'Hôpital's Rule.

Step 6: Apply L'Hôpital's Rule (or Taylor Series). Applying L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 0} {{\frac{d}{dx}(1 - \cos x)} \over {\frac{d}{dx}({x^2})}} = \mathop {\lim }\limits_{x \to 0} {{\sin x} \over {2x}}$ This is still of the form $\frac{0}{0}$, so we apply L'Hôpital's Rule again: $L = \mathop {\lim }\limits_{x \to 0} {{\frac{d}{dx}(\sin x)} \over {\frac{d}{dx}(2x)}} = \mathop {\lim }\limits_{x \to 0} {{\cos x} \over 2}$ Substituting $x=0$: $L = {{\cos 0} \over 2} = {1 \over 2}$

Alternatively, using the Taylor series expansion for $\cos x$ around $x=0$, $\cos x = 1 - \frac{x^2}{2} + O(x^4)$: $1 - \cos x = 1 - \left(1 - \frac{x^2}{2} + O(x^4)\right) = \frac{x^2}{2} + O(x^4)$ Then the limit becomes: $L = \mathop {\lim }\limits_{x \to 0} {{\frac{x^2}{2} + O(x^4)} \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} \left({1 \over 2} + O(x^2)\right) = {1 \over 2}$

3. Common Mistakes & Tips

• Squaring both sides: Be cautious about introducing extraneous solutions. Always check the derived function against the original conditions. In this case, $f(x) \ge 0$ helped select the correct form of the solution.
• Differentiating integrals: Ensure correct application of Leibniz's Rule, especially when the limits of integration are functions of the variable of differentiation.
• Indeterminate forms: When evaluating limits, always check if the form is indeterminate ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule.

4. Summary

The problem requires transforming an integral equation into a differential equation by differentiating both sides using Leibniz's Rule. Solving the resulting separable differential equation with the given initial condition yields $f(x) = \sin(x)$. Finally, the limit is evaluated by integrating $f(t)$ and applying L'Hôpital's Rule or Taylor series expansion for the indeterminate form.

The final answer is $\frac{1}{2}$, which corresponds to option (D).

The final answer is $\boxed{\frac{1}{2}}$.