A doubt about the central limit theorem I have a doubt about the central limit theorem.
Let $(X_n)$ be an IID sequence, each $X_n$ distributed as $X$ where
$\mathbb{E}[X]=0$, $\sigma^2 = Var(X) < \infty$.
Define $S_n = X_1 + \dots + X_n$, and set
$G_n = \frac{S_n}{\sigma\sqrt{n}}$
Now, the theorem tells us that
$\lim_{n \to \infty}\mathbb{P}(G_n \le x) = \Phi(x)$ where $\Phi(x)$ is the distribution function of the standard normal distribution.
Now is it FORMALLY correct to say, for example, that
$\lim_{n \to \infty}\mathbb{P}(G_n \le 1+\frac{1}{n}) = \Phi(\lim_{n \to \infty}(1+\frac{1}{n})) = \Phi(1) \simeq 0.8413$?
Thanks to who will solve my (perhaps trivial) doubt.
 A: No, it is not formally correct because $1+1/n$ is a sequence of numbers which depends on $n$, whereas the CLT as stated only holds for fixed real numbers $x$.
However, it is still true that $\lim_{n \to \infty} P(G_n \leq 1+1/n) = \Phi(1)$. In order to prove this, one only needs to use the fact that the functions $F_n(x):=P(G_n \leq x)$ are increasing, together with continuity of the limit $\Phi$. Indeed, for $N \leq n$, one has that $F_n(1+1/n) \leq F_n(1+1/N)$, and thus for any fixed $N$ $$\limsup_{n\to \infty}F_n(1+1/n) \leq \limsup_{n \to \infty} F_n(1+1/N) = \Phi(1+1/N)$$ Next, we may let $N \to \infty$ on both sides, and since the LHS does not depend on $N$, one sees by continuity of $\Phi$ that $$\limsup_{n \to \infty}F_n(1+1/n) \leq \lim_{N\to \infty} \Phi(1+1/N) = \Phi(1)$$ 
On the other hand, one clearly has that $F_n(1+1/n) \geq F_n(1)$ fo every $n$, and therefore we see that $$\liminf_{N \to \infty} F_n(1+1/n) \geq \liminf_{n \to \infty} F_n(1) = \Phi(1)$$ which (together with the preceding expression) gives the desired result.
In fact, because of this question, it is actually true that one has uniform convergence of the cdf's $F_n$ to $\Phi$, which means that $F_n(x_n) \to \Phi(x)$ whenever $x_n \to x$ (and the proof uses the same ideas).
