Let $X_n$ be a sequence of independent random variables with $X_n$ having the probability density function of $\chi ^2_{(n)}$(chi-square distribution with $n$ degrees of freedom). Then find $\lim_{n \rightarrow \infty} [P(X_n>\frac{3}{4}n)+P(X_n>n+2\sqrt{2n})]$
Now, we know by Central Limit theorem $\lim_{n \rightarrow \infty} \frac{X_n-n}{\sqrt{2n}} \rightarrow N(0,1)$ Thus, $P(X_n>n+2\sqrt{2n})=1- \Phi(2)$. But, I cannot find $\lim_{n \rightarrow \infty} P(X_n>\frac{3}{4}n)$.Please Help!
$$ \mathbb P\left(X_n>\frac34 n\right) = \mathbb P\left(\frac{X_n-n}{\sqrt{2n}}>\frac{\frac34 n - n}{\sqrt{2n}}\right)=\mathbb P\left(\frac{X_n-n}{\sqrt{2n}}>-\frac{\sqrt{n}}{4\sqrt{2}}\right)\to 1 $$ as $n\to\infty$ since $-\frac{\sqrt{n}}{4\sqrt{2}}\to-\infty$ and probability that limiting standard normally distributed r.v. is greater than $-\infty$ is $1$.