Let

$$L(x)=Q\left(\frac{x}{2},\frac{a}{a+f(x)/\sqrt{x}}Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)}\right)\right)$$

where $Q(s,x)=\frac{\Gamma(s,x)}{\Gamma(s)}$ is the upper incomplete gamma function $\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$ regularized by gamma function $\Gamma(s)$, $Q^{-1}(s,z)$ is the solution for $x$ in $z=Q(s,x)$, $a>0$ is a constant, $\frac{1}{2}<b<1$ is also a constant, and $g(x)=\omega(x)$ is a function that is asymptotically greater than $x$.

I am interested in the behavior of the limit $\lim_{x\rightarrow\infty}L(x)$ for varying asymptotics of $f(x)$. Specifically, numerical evaluations seem to indicate that when $f(x)=\omega(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=0$ and when $f(x)=\mathcal{O}(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=1$. However, I am having trouble proving this analytically (my usual Taylor expansion tricks don't seem to work here). Any help?

Let

$$L(x)=Q\left(\frac{x}{2},\frac{a}{a+f(x)/\sqrt{x}}Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)}\right)\right)$$

where $Q(s,x)=\frac{\Gamma(s,x)}{\Gamma(s)}$ is the upper incomplete gamma function $\Gamma(s,x)=\int_x^\infty t^{s-1}e^{-t}dt$ regularized by gamma function $\Gamma(s)$, $Q^{-1}(s,z)$ is the solution for $x$ in $z=Q(s,x)$, $a>0$ is a constant, $\frac{1}{2}<b<1$ is also a constant, and $g(x)=\omega(x)$ is a function that is asymptotically greater than $x$.

I am interested in the behavior of the limit $\lim_{x\rightarrow\infty}L(x)$ for varying asymptotics of $f(x)$. Specifically, numerical evaluations seem to indicate that when $f(x)=\omega(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=0$ and when $f(x)=\mathcal{O}(\sqrt{\log(x)})$, $\lim_{x\rightarrow\infty}L(x)=1$. However, I am having trouble proving this analytically (my usual Taylor expansion tricks don't seem to work here). Any help?