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Step-by-Step Solution
Step 1: Understand the Given Condition
We are given a non-negative function $f$ on $[0,1]$ such that
$$
\int_{0}^{x} \sqrt{1 - (f'(t))^2}\,dt \;=\; \int_{0}^{x} f(t)\,dt,\quad 0 \le x \le 1,
$$
and $f(0)=0.$ Our goal is to evaluate
$$
\lim_{x \to 0} \frac{1}{x^2}\,\int_{0}^{x} f(t)\,dt.
$$
Step 2: Differentiate Both Sides
By the Fundamental Theorem of Calculus, differentiate both integrals with respect to $x$:
$$
\frac{d}{dx} \left(\int_{0}^{x} \sqrt{1 - (f'(t))^2}\,dt \right)
\;=\; \frac{d}{dx} \left(\int_{0}^{x} f(t)\,dt \right).
$$
Hence,
$$
\sqrt{1 - (f'(x))^2} \;=\; f(x).
$$
Step 3: Rewrite and Solve the Resulting Equation
Square both sides to get:
$$
1 - (f'(x))^2 \;=\; f(x)^2
\;\;\Longrightarrow\;\; (f'(x))^2 + f(x)^2 \;=\; 1.
$$
Taking the positive branch (since $f$ is non-negative and the square root is also non-negative), we write:
$$
f'(x) \;=\; \sqrt{1 - f(x)^2}.
$$
Step 4: Separate Variables and Integrate
Separate the variables:
$$
\frac{df}{\sqrt{1 - f^2}} \;=\; dx.
$$
The integral of the left side is $\sin^{-1}(f)$, so we have
$$
\sin^{-1}\bigl(f(x)\bigr) \;=\; x + C.
$$
Using the condition $f(0) = 0$, we get
$$
\sin^{-1}\bigl(f(0)\bigr) \;=\; \sin^{-1}(0) \;=\; 0,
$$
which implies $C=0$. Thus,
$$
\sin^{-1}\bigl(f(x)\bigr) \;=\; x
\;\;\Longrightarrow\;\;
f(x) \;=\; \sin x.
$$
Step 5: Evaluate the Required Limit
We now need to compute
$$
\lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} f(t)\,dt
\;=\;
\lim_{x \to 0} \frac{1}{x^2} \int_{0}^{x} \sin t\,dt.
$$
The integral of $\sin t$ over $[0,x]$ is
$$
\int_{0}^{x} \sin t \, dt
\;=\;
\bigl[-\cos t \bigr]_{0}^{x}
\;=\;
1 - \cos x.
$$
Therefore, our limit becomes
$$
\lim_{x \to 0} \frac{1 - \cos x}{x^2}.
$$
Step 6: Simplify the Limit
Recall the standard limit result:
$$
\lim_{x \to 0} \frac{1 - \cos x}{x^2} \;=\; \frac{1}{2}.
$$
Hence,
$$
\lim_{x \to 0} \frac{1}{x^2}\,\int_{0}^{x} f(t)\,dt
\;=\;
\frac{1}{2}.
$$
Final Answer
The desired limit equals $\frac{1}{2}$.
