Can a normal approximation assumption justify itself? I am learning elementary statistics.
I found an exercise, which asks to compute the desired sample size for some interval for standard error.
The solution, in class slides, first assumes the sample size to be computed is big enough to be approximated by a normal distribution (central limit theorem), and uses it to get an answer that matches the assumption — the sample size is big enough. 
Is this is a sound answer?  I don't think so; it may be a wrong assumption first, and that result deduced from it is not qualified to do any validation, is it? 
If it is not the case, how can we prove that assumption first? Or put another way,if I want to compute a desirable sample size, is it that you can never allow using the central limit theorem on that answer/question?
*I have no teacher to ask because I'm self-studying.
edit I think the post is not quite clear, the C.T.L. is used upon a random variable that is the sum of all samples to be estimated.
 A: The normal approximation is not some 'on/off' thing - it improves slowly as sample size increases and gets slowly worse as it decreases.
Which is to say, when you assume it, though it may lead you to a wrong sample size, if the sample size it leads you to is large enough*, the normal approximation at any nearby sample size should also be reasonable. If the sample size it leads you to is small, the normal approximation may be poor. 
* I don't know that you would yet have a basis to conclude that in general; simulation will generally be a useful way to check
That is, if you're a little bit conservative about invoking the calculated sample size as a reason to use the normal-assumption value you calculated, it will usually work quite well.
When in doubt, you can sometimes use algebra or simulation to check how things work at the sample size you calculate; it depends on the exact details of the situation.
To my mind some of the other assumptions are more likely to cause you problems than that one.
A: In real life, although the actual distribution is unknown, there are some hints if the normal approximation ("the $t$-statistic is $t_n$-distributed as if the data were normally distributed") is reasonable at moderate sample sizes or not.
I assume you had to find the sample size $n$ for a $1-\alpha$-confidence interval for the parameter $E(X_i), i=1,\ldots,n$ that had to be $a\sqrt{E(X_i -E(X_i))^2}$ large.
If the true distribution is not much apart from the normal distribution (in particular quite symmetric) and $\alpha$ is large enough, you can proceed the way your textbook solution suggested. In this case the normal approximation is pretty fast. Just simulate some distributions and watch how far the actual coverage of $\mu$ is missed by the $t$-based CI. This is also statistician's experience.
If the true distribution is quite right-skewed, you can take the Berry-Esseen-Bound to adjust for possible liberalness due to nonormality and small sample size. Your conservative (i.e. $n$ may be too large, true coverage probability may be larger than $1-\alpha$) solution would then be the minimum integer $n$ solving
$$ 2 t_{n-1,1-\frac{\alpha}{2}+\frac{C\rho}{\sigma^3 n}}\leq a\sigma$$
A: It's only a problem if you got a sample size smaller than what is required to assume for normality.  
