Let f be a non - negative function in $[0,1]$ and twice differentiable in $(0,1)$. If $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f$, $0 \leq x \leq 1$ and $\mathrm{f}(0)=0$, then $\lim _{x \rightarrow \infty} \frac{1}{x^2} \int_0^x f(t) d t$ :
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