How can we get a normal distribution as $n \to \infty$ if the range of values of our random variable is bounded? Let's say we have a random variable with a range of values bounded by $a$ and $b$, where $a$ is the minimum value and $b$ the maximum value.
I was told that as $n \to \infty$, where $n$ is our sample size, the sampling distribution of our sample means is a normal distribution. That is, as we increase $n$ we get closer and closer to a normal distribution, but the actual limit as $n \to \infty$ is equal to a normal distribution.
However, isn't part of the definition of the normal distribution that it has to extend from $- \infty$ to $\infty$?
If the max of our range is $b$, then the maximum sample mean (regardless of sample size) is going to be equal to $b$, and the minimum sample mean equal to $a$. 
So it seems to me that even if we take the limit as $n$ approaches infinity, our distribution is not an actual normal distribution, because it is bounded by $a$ and $b$.
What am I missing?
 A: If you're referring to a central limit theorem, note that one proper way to to write it out is
$\left( \frac{\bar x - \mu} {\sigma} \right)  \sqrt n \rightarrow_d N(0,1)$
under normal conditions ($\mu, \sigma$ being the mean and standard deviation of $x_i$).
With this formal definition, you can see right away that the left hand side can take on values for any finite range given a large enough $n$. 
To help connect to the for informal idea that "a mean approaches a normal distribution for large $n$", we need to realize that "approaches a normal distribution" means that the CDF's get arbitrarily close to a normal distribution as $n$ gets large. But as $n$ gets large, the standard deviation of this approximate distribution shrinks, so the probability of an extreme tail of the approximating normal also goes to 0.
For example, suppose $X_i \sim \text{Bern}(p = 0.5)$. Then you could use the informal approximation to say that 
$\bar X \dot \sim N\left(p, \frac{p(1-p)}{n}\right)$
So while it is true that for any finite $n$, 
$P\left(N\left(p, \frac{p(1-p)}{n}\right) < 0\right) >0$
(implying the approximation is clearly never perfect), as $n \rightarrow \infty$, 
$P\left(N\left(p, \frac{p(1-p)}{n}\right) < 0\right) \rightarrow 0$
So that discrepancy between the actual distribution and approximate distribution is disappearing, as is supposed to happen with approximations. 
A: Here is what you are missing. The asymptotic distribution is not of $\bar{X}_n$ (the sample mean), but of $\sqrt{n}(\bar{X}_n - \theta)$, where $\theta$ is the mean of $X$.
Let $X_1, X_2, \dots$ be iid random variables such that $a < X_i <b$ and $X_i$ has mean $\theta$ and variance $\sigma^2$. Thus $X_i$ has bounded support. The CLT says that
$$\sqrt{n}(\bar{X}_n - \theta) \overset{d}{\to} N(0, \sigma^2), $$
where $\bar{X}_n$ is the sample mean. Now
\begin{align*}
a < &X_i <b\\
a < & \bar{X}_n <b\\
a-\theta < &\bar{X}_n - \theta < b - \theta\\
\sqrt{n}(a - \theta) < & \sqrt{n}(\bar{X}_n - \theta) < \sqrt{n}(b - \theta).\\
\end{align*}
As $n \to \infty$, the lower bound and the upper bound tend to $-\infty$ and $\infty$ respectively, and thus as $n \to \infty$ the support of $\sqrt{n}(\bar{X}_n - \theta)$ is exactly the whole real line.
Whenever we use the CLT in practice, we say $\bar{X}_n \approx N(\theta, \sigma^2/n)$, and this will always be an approximation.

EDIT: I think part of the confusion is from the misinterpretation of the Central Limit Theorem. You are correct that the sampling distribution of the sample mean is 
$$\bar{X}_n \approx N(\theta, \sigma^2/n). $$
However, the sampling distribution is a finite sample property. Like you said, we want to let $n \to \infty$; once we do that the $\approx$ sign will be an exact result. However, if we let $n \to \infty$, we can no longer have an $n$ on the right hand side (since $n$ is now $\infty$). So the following statement is incorrect
$$ \bar{X}_n \overset{d}{\to} N(\theta, \sigma^2/n) \text{ as } n \to \infty.$$
[Here $\overset{d}{\to}$ stands for convergence in terms of distribution]. We want to write the result down accurately, so the $n$ is not on the right hand side. Here we now use properties of random variables to get
$$ \sqrt{n}(\bar{X}_n - \theta) \overset{d}{\to} N(0, \sigma^2)$$
To see how the algebra works out, look at the answer here.
