In the following proof I understand that we have taken power n because for sum of variables we take product of characterstic functions. Intuitively I understand why $n \to \infty$ is important but what is the mathematical reason for the condition?
Let $X_1, X_2, \cdots$ be iid random variables each having mean $\mu$ and finite non zero variance $\sigma^2$. Then
$\lim_{n \to \infty} P(\frac{S_n-n \mu}{\sigma \sqrt{n}} \leq x)=\Phi(x)$, $-\infty<x<\infty$
Proof:
$S^*_n=\frac{S_n-n\mu}{\sqrt{n}\sigma}$
Then for t fixed and sufficiently large n,
$\phi_{S_n^*}(t)=e^{-in\mu t/\sigma \sqrt{n}}\phi_{S_n}(t/\sigma \sqrt{n})=e^{-in\mu t/\sigma \sqrt{n}}(\phi_{X_1}(t/\sigma \sqrt{n}))^n$
Well, the CLT used in practice is about getting approximations to distributions. But to get a limit result, we need simplifications. While we are in practice interested in the distribution of an average for large, but finite, $n$, without letting $n \rightarrow \infty$ we cannot say much about the distribution, we cannot say anything in general. So the limit is necessary to get a result at all!
But a limit result is a qualitative result. It says that eventually, for sample size $n$ "large enough" we get very close to a normal distribution. But it dosn't say anything about "how large $n$ must be" to get "close enough". (neither did we define what we mean by "close enough") Such quantitative results might be useful, but they are more difficult to get, to prove, and above all, to use.