Limit of chi square distribution. Question in my exercise is written as $$\lim_{n \to \infty}\bigg(\dfrac{1}{2^{\frac{n}{2}}\Gamma{\frac{n}{2}}}\int_{n+\sqrt{2n}}^{\infty}e^{\frac{-t}{2}}t^{\frac{n}{2}-1}dt\bigg)$$ equals : 
$(A)=.5$
$(B)=0$
$(C)=.0228$
$(D)=.1587$
As sample size increase chi square approaches normal distribution.(I am not sure if i wrote this statement correct so please correct me and give me little explanation on that). Using this fact i calculated $P(X>n-\sqrt{2n}) = \Phi(-1) = .1587$. Did i do everything correct using this intuition ? 
 A: Remark. The question is changed after posting this answer. The lower bound of the integral was $n-\sqrt[] {2n}$ first. The idea remains the same. 
Your idea to use the Central Limit Theorem is good. First note that the sum of chi distributed random variables is again chi distributed.  Let $U_i\sim \chi_1^2$  i.i.d. then $\sum_{i=1}^n U_i\sim \chi_n^2 $. What you want to find is:
\begin{align}
\lim_{n\to\infty} \mathbb P\left(\sum_{i=1}^n U_i>n-\sqrt[]{2n}\right) = \lim_{n\to\infty} \frac{1}{2^{n/2}\Gamma(n/2)}\int^\infty_{n-\sqrt[]{2n}}e^{-t/2}t^{\frac{n}{2}-1}\,dt
\end{align}
The LHS is:
\begin{align}
\lim_{n\to \infty}\mathbb P\left(\frac{\sum_{i=1}^n U_i-n}{\sqrt[]{2n}}>-1\right)&=\lim_{n\to \infty}\mathbb P\left(\frac{\sum_{i=1}^n U_i-n\mathbb E[U_1]}{\sqrt[]{n\operatorname{Var}(U_1)}}>-1\right)\\&\stackrel{CLT}{=} 1-\Phi(-1)
\end{align}
where $\Phi(\cdot)$ is the CDF of standard normal distribution. But $1-\Phi(-1)=\Phi(1)\approx 0.8413447$ which is not one of the options. Strange..
We conclude anyway that:

\begin{align}
\lim_{n\to\infty} \frac{1}{2^{n/2}\Gamma(n/2)}\int^\infty_{n-\sqrt[]{2n}}e^{-t/2}t^{\frac{n}{2}-1}\,dt=\Phi(1)\approx 0.8413447
\end{align}

