Limit involving regularized gamma function and its inverse 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?
 A: Below I give a sketch of a proof.
First an approximation for the inverse incomplete gamma function, $Q^{-1}$, is needed.
Henceforth I assume that
$$
g(x)\sim \gamma x^s 
$$
for large $x$ with $s>0$. Then ($0<b<1$)
$$
1-b^{1/g(x)}\sim (\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right),
$$
which is small for large $x$.
Now we use a formula due to Tricomi, which can be found here
$$
Q(\alpha+1,\alpha+\sqrt{2 \alpha} y) = \frac{1}{2} \text{erfc}(y),
$$
for $\alpha\rightarrow \infty$.
Setting $\alpha=x/2$ one gets for large $x$ (dropping constant and lower terms) the following approximation
$$
Q\left(\frac{x}{2},\frac{x}{2}\left(1+\frac{2 y}{\sqrt{x}}\right)\right) \sim \frac{1}{2} \text{erfc}(y) .
$$
Now we choose $y$ such that the right side is approximately equal to $(\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right)$. With the help of the expansion of the inverse erfc, $\text{inverfc}$ (see e.g.  here .)
$$
\text{inverfc} (y) \sim \frac{1}{\sqrt{2}} \sqrt{-\ln\left(\pi y^2 \ln \frac{1}{y}\right)} .
$$
Again using the fact that $x$ is large, we find $y\sim \sqrt{s} \sqrt{\ln x}$, keeping only the highest order in $\ln x$.
This gives
$$
Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)} \right) \sim \frac{x}{2}\left(1+2\sqrt{ s}\sqrt{ \frac{\ln x}{x}}\right),
$$
independent of $\gamma$ and $b$.
Insertion and once more dropping terms of lower order in $x$ (assuming $f(x)/\sqrt{x}\rightarrow 0$ for $x\rightarrow \infty$)
$$
L(x) \sim Q\left(\frac{x}{2},\frac{x}{2}\left[1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right)\right]\right).
$$
Another asymptotic formula for the incomplete gamma functions comes into play (see, e.g.,   here):
$$
\Gamma(z,\lambda z) \sim (\lambda z)^z e^{-\lambda z}\frac{1}{z(\lambda - 1)},
$$
For this formula to be valid $\lambda > 1$.
Using the above and Stirling's formula for the $\Gamma(z)$ with $z=x/2$ and setting
$$
\lambda = \lambda(x) := 1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right).
$$
after a lot of cancellations the result is
$$
L(x) \sim (2 \sqrt{s \ln x}- f(x)/a)^{-1}\rightarrow 0
$$
for $x \rightarrow \infty$.
Note that
$$
\lambda(x)^{x/2} \sim e^{x(\lambda(x)-1)/2},
$$
which cancels exactly with the exponential term in the approximation, thus leaving only the $\sqrt{x}/(\lambda(x)-1)$.
If we have $\lambda = 1$, i.e. $f(x) = 2 a \sqrt{s \ln x}$, another formula might be used (see, e.g. here)
$$
\Gamma(z,z) \sim \sqrt{\frac{\pi}{2}}  z^{z-\frac{1}{2}}e^{-z} .
$$
With this I find
$$
L(x)\sim Q \left(\frac{x}{2},\frac{x}{2}\right)\sim \frac{1}{2},
$$
which is in disagreement with the numerical results. One has instead to take into account next order terms.
