Definite Integration

Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If \(\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} }\), \(0 \le x \le 1\) and f(0) = 0, then \(\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}\) :