How to find asymptotics of integrand?

Question

Let $ f \in C ([0, \infty)) $ be s. t. $$f(x) \int_0^x f(t)^2 dt \to 1, x \to \infty.$$ How to prove that $f(x) \sim \left( \frac 1 {3x} \right)^{1/3} $ as $x \to \infty?$

Answer 1

Define $$ F(x)=\int_0^xf(t)^2\,\mathrm{d}t\tag{1} $$ For any $\epsilon\gt0$ there is an $M$, so that if $x\ge M$, $$ 1-\epsilon\le f(x)F(x)\le1+\epsilon\tag{2} $$ Squaring yields $$ (1-\epsilon)^2 \le F(x)^2\frac{\mathrm{d}}{\mathrm{d}x}F(x) \le(1+\epsilon)^2\tag{3} $$ Thus, $$ 3(1-\epsilon)^2 \le\frac{\mathrm{d}}{\mathrm{d}x}F(x)^3 \le3(1+\epsilon)^2\tag{4} $$ and so, $$ \left(3(1-\epsilon)^2x+C\right)^{1/3}\le F(x)\le\left(3(1+\epsilon)^2x+C\right)^{1/3}\tag{5} $$ and finally, $$ \frac{1-\epsilon}{\left(3(1+\epsilon)^2x+C\right)^{1/3}} \le f(x) \le\frac{1+\epsilon}{\left(3(1-\epsilon)^2x+C\right)^{1/3}}\tag{6} $$ Therefore, by the Squeeze Theorem, $$ \lim_{x\to\infty}f(x)(3x)^{1/3}=1\tag{7} $$