Here is the problem, I want to calculate the limit of this function when $x\to\infty$.
$$\lim_{x\to\infty}\displaystyle\frac{\gamma(x,-c\sqrt{x}+x)}{\Gamma(x,c\sqrt{x}+x)},$$ where $\gamma(a,b)$ and $\Gamma(a,b)$ are the lower and upper incomplet gamma function, respectively.
$$\gamma(a,b)=\int_{0}^bt^{a-1}e^{-t}dt,\quad\text{and}\quad\Gamma(a,b)=\int_{b}^\infty t^{a-1}e^{-t}dt.$$
Numerically, I think the answer is 1. I tried several approximations for incomplete gamma function, but it did not work out :(
1st Solution. Let $(X_t)_{t\geq 0}$ be the gamma process. We know that each $X_t$ has the gamma distribution:
$$ \mathbf{P}(X_t \in \mathrm{d}x) = \frac{x^{t-1}e^{-x}}{\Gamma(t)} \, \mathrm{d}x. $$
Then by CLT, we find that $Z_t := \frac{X_t - t}{\sqrt{t}}$ converges in distribution to the standard normal distribution. Hence we get
\begin{align*} \frac{\gamma(t, t-c\sqrt{t})}{\Gamma(t, t+c\sqrt{t})} &= \frac{\mathbf{P}(X_t \leq t-c\sqrt{t})}{\mathbf{P}(X_t \geq t+c\sqrt{t})} = \frac{\mathbf{P}(Z_t \leq -c)}{\mathbf{P}(Z_t \geq c)} \\ &\to \frac{\mathbf{P}(Z \leq -c)}{\mathbf{P}(Z \geq c)} = 1, \end{align*}
where $Z$ is a standard normal variable.
2nd Solution. For $\gamma$, we substitute $t = x - s\sqrt{x}$. Then
\begin{align*} \frac{\gamma(x, x-c\sqrt{x})}{\Gamma(x)} &= \frac{x^{x-1/2}e^{-x}}{\Gamma(x)} \int_{c}^{\sqrt{x}} \left(1 - \frac{s}{\sqrt{x}}\right)^{x-1} e^{s\sqrt{x}} \, \mathrm{d}s. \end{align*}
Now, using the inequality $\log(1+x) \leq x - \frac{x^2}{2(1+x_+)}$ that is valid for any $x > -1$, we can check that the integrand is dominated by an integrable function on $[0, \infty)$. Also, the prefactor $\frac{x^{x-1/2}e^{-x}}{\Gamma(x)}$ converges to $\frac{1}{\sqrt{2\pi}}$ by the Stirling's approximation. So, by the dominated convergence theorem,
\begin{align*} \lim_{x \to \infty} \frac{\gamma(x, x-c\sqrt{x})}{\Gamma(x)} &= \frac{1}{\sqrt{2\pi}} \int_{c}^{\infty} e^{-s^2/2} \, \mathrm{d}s. \end{align*}
Similarly, for $\Gamma$ we may substitute $t = x + s\sqrt{x}$ to get
\begin{align*} \lim_{x\to\infty} \frac{\Gamma(x, x+c\sqrt{x})}{\Gamma(x)} &= \lim_{x\to\infty} \frac{x^{x-1/2}e^{-x}}{\Gamma(x)} \int_{c}^{\infty} \left(1 + \frac{s}{\sqrt{x}}\right)^{x-1} e^{-s\sqrt{x}} \, \mathrm{d}s \\ &= \frac{1}{\sqrt{2\pi}} \int_{c}^{\infty} e^{-s^2/2} \, \mathrm{d}s. \end{align*}
Therefore the deisred limit is $1$.