When to use Central Limit Theorem or Cramers Theorem In for example this paper the authors say

The central limit theorem provides an estimate of the probability
  \begin{align}
P\left( \frac{\sum_{i=1}^n X_i - n\mu}{\sigma \sqrt{n}} > x \right)
\end{align}
  ... the CLT estimates the probability of $O(\sqrt{n})$ deviations from the mean of the sum of random variables ... On the other hand, large deviations of the order of the mean itself, i.e., $O(n)$ deviations, is the subject of this section [Cramer-Chernoff Theorem].

It is not clear to my why the CTL can't be used to calculate large deviations. Following the answer of my previous question for large $n$, the CTL tells me, that the mean is approximately normally distributed as
$$P\left(|\sum_{i=1}^n X_i - n\mu| \geq x\right) \approx 2\Phi\left(-\frac{x \sqrt{n}}{\sigma}\right)$$ 
Why (and in which cases) should Cramers theorem be used if $x$ is large and not the CTL?

Large deviation gives you an estimate of the probabilities in the non-typical regime, whereas CLT gives you an estimate of the probabilities in the typical regime. Suppose $X_i$ have finite variance and are a.s. positive. Then CLT gives,
$$
|P(\frac{\sum_i X_i  - n\mu}{\sigma \sqrt{n}} \ge x) - \Phi(-x)| \to 0.
$$
But CLT does'nt say anything if you let $x$ grow with $n$ as well. Say you want to know $P(\sum_i X_i - n\mu >\mu n)$ and $X_i$'s are all positive a.s. Then replacing $x$ by $\sqrt{n} \mu/\sigma$ would give you an approximation of this probability as $e^{-cn}$ from some $c>0$. But this would not be correct. Lets say, $X_i$ have a very heavy tail, $P(X_i > n) \ge n^{-4}$. Then if $X_1 >2n\mu$ then $\sum_{i \le n}X_i > 2n\mu$ and hence
$$
P(|\sum_{i \le n}X_i - n\mu| > n\mu) \ge P(\sum_{i \le n}X_i > 2n\mu)  \ge P(X_1 > 2n\mu) \ge (2n\mu)^{-4}
$$
and you do not get an exponential bound, but a polynomial bound.
