Limit $ \lim\limits_{n\to\infty}\Bigl[\frac{1}{2^{n/2}\Gamma(n/2)} \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt\Bigr]$ Find $ \lim\limits_{n\to\infty}\left[\dfrac{1}{2^{n/2}\Gamma(n/2)}
\displaystyle \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt\right]$
This looks like the p.d.f. of a chi-square distribution. With limit $n\to \infty$, this should be a normal distribution. So, should this be 1 (like the area under the curve) ?
What does $n-\sqrt{2n}$ mean here ?
Please help.
 A: Set $m=n/2$, so that you want to compute
$$
\lim\limits_{m\to\infty}\left[\dfrac{1}{2^{m}\Gamma(m)}
\displaystyle \int_{2m-2\sqrt{m}}^{\infty} t^{m-1}e^{\frac{-t}{2}}\,dt\right]
=\lim\limits_{m\to\infty}\mathbb P(X_m\ge 2m-2\sqrt m),
$$
where $X_m\sim\Gamma(m,1/2)$. Now, use the fact that $X_m$ has the same distribution as $\sum_{k=1}^mE_k$, where $E_k\sim\mathrm{Exp}(1/2)$ are i.i.d. random variables.
Then, 
$$
\mathbb P(X_m\ge 2m-2\sqrt m)
=\mathbb P\left(\sum_{k=1}^mE_k\ge 2m-2\sqrt m\right)
=\mathbb P\left(\sqrt m\left(\frac1m\sum_{k=1}^mE_k-2\right)\ge-2\right),
$$
and by the CLT, $\sqrt m\left(\frac1m\sum_{k=1}^mE_k-2\right)\xrightarrow[m\to\infty]{}\mathcal N(0,4)$ in distribution, so
$$
\mathbb P(X_m\ge 2m-2\sqrt m)
\xrightarrow[m\to\infty]{}\mathbb P\left(N\ge-1\right)\left(=\frac1{\sqrt{2\pi}}\int_{-1}^\infty\exp\left(-x^2/2\right)\,\mathrm dx\right),
$$
where $N\sim\mathcal N(0,1)$.
One has $\mathbb P\left(N\ge-1\right)\approx0.841345$, which is in line with Claude Leibovici's answer.
A: Being totally ignorant in the area of probability distribution functions (just as in many other areas), what I am writing could be totally off-topic. If this would be the case, please forgive me.
First the antiderivative
$$ \int t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=-2^{n/2} \Gamma \left(\frac{n}{2},\frac{t}{2}\right)$$ So, the definite integral
$$ \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=2^{n/2} \Gamma \left(\frac{n}{2},\frac{1}{2} \left(n-\sqrt{2n}\right)\right)$$ So,  the expression $$\dfrac{1}{2^{n/2}\Gamma(n/2)}
\displaystyle \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=\frac{\Gamma \left(\frac{n}{2},\frac{1}{2} \left(n-\sqrt{2n}
  \right)\right)}{\Gamma \left(\frac{n}{2}\right)}$$ I was hoping to find appropriate expansions to be useful for the evaluation of the limit but I have been unsuccessful and I apologize for that.
Numerical evaluations for very large values of $n$ (up to $5\times 10^6$) show values close to $0.841345$.
Added after Ian's comments and edited answer
Taking into account the new results given by Ian (to whole credit has to be given), then the limit is simply $$\frac{1}{2} \left(1+\text{erf}\left(\frac{1}{\sqrt{2}}\right)\right)\approx 0.841345$$
