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Let $\gamma(s,x)$ and $\Gamma(s,x)$ denote the lower and upper incomplete gamma function, respectively.

My question concerns the existence of the following limit $$ \lim_{x\to\infty}\left(\frac{\gamma(2x+1,2bx)}{a+\Gamma(2x+1,2bx)}\right)^{\frac{1}{x}}, $$ where $a,b>0$ are (strictly) positive real numbers.

Note 1. Simulations seem to suggest that the above limit exists (for any choice of $a,b$, as above) and it is in general different from zero.

Note 2. Seemingly related questions: Q1, Q2.

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    For starters, notice that the choice for *a* doesn't really matter. Letting $t=2x,$ it all then boils down to finding the square of the limit $\displaystyle\lim_{t\to\infty}\sqrt[\Large^t]{\frac{\Gamma(t)}{\Gamma(t,bt)}-1}.$ – 2017-02-23
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    Which itself is nothing other than $\displaystyle\lim_{t\to\infty}\sqrt[\Large^t]{\frac{\Gamma(t)}{\Gamma(t,bt)}}.$ – 2017-02-23
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    The limit is simply $~\dfrac{e^{b-1}}b.$ – 2017-02-23
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    And, as I've already said earlier, your limit is its square. – 2017-02-23
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    Note, however, that the above results only work for $b\ge1.$ – 2017-02-23
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    @Lucian: Thanks for the comments! I would really appreciate if you could add some more details, concerning in particular your last three comments. (Perhaps you could collect this in an answer, so that I can accept it). – 2017-02-23
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    @Lucian: After having thought about this for a while, I'm still missing something: How do you show that for $b\geq 1$ $$\lim_{t\to \infty} \left(\frac{\Gamma(t)}{\Gamma(t,bt)}\right)^{\frac{1}{t}}=\frac{e^{b-1}}{b}\ \ ?$$ (And what about the case $0< b<1$?) – 2017-03-03

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