I know the coefficients of the continued fraction representation of a function that diverges like $O(\sqrt{x})$, where the $a_k$ depend on $x$,

$ f(x)=1+a_0/\left(1+\underset{k=1}{\overset{\infty }{K}}\frac{-a_k}{a_k+1}\right)=1+\frac{\displaystyle{a_0}}{\displaystyle{1-\frac{a_1}{(1+a_1)-\displaystyle\frac{a_2}{(1+a_2)-\displaystyle\frac{a_3}{\dots}}}}} $

Is there a way of computing $\lim_{x\rightarrow\infty}\frac{f(x)}{\sqrt{x}}$? Is should be $\frac{3}{4}\sqrt{\frac{\pi}{2}}$.

EDIT: For completeness, $a_k=\frac{\left(k+\frac{1}{2}\right) (k-2 x+1)^3}{(k+1) \left(k-2 x+\frac{1}{2}\right)^3}$.


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