What is the distribution of sample means of a Cauchy distribution? Typically when one takes random sample averages of a distribution (with sample size greater than 30) one obtains a normal distribution centering around the mean value. However, I heard that the Cauchy distribution has no mean value. What distribution does one obtain then when obtaining sample means of the Cauchy distribution?
Basically for a Cauchy distribution $\mu_x$ is undefined so what is $\mu_{\bar{x}}$ and what is the distribution of $\bar{x}$?
 A: 
Typically when one takes random sample averages of a distribution (with sample size greater than 30) one obtains a normal distribution centering around the mean value.

Not exactly. You're thinking of the central limit theorem, which states that given a sequence $X_n$ of IID random variables with finite variance (which itself implies a finite mean $μ$), the expression $\sqrt{n}[(X_1 + X_2 + \cdots + X_n)/n - μ]$ converges in distribution to a normal distribution as $n$ goes to infinity. There is no guarantee that the sample mean of any finite subset of the variables will be normally distributed.

However, I heard that the Cauchy distribution has no mean value. What distribution does one obtain then when obtaining sample means of the Cauchy distribution?

Like GeoMatt22 said, the sample means will be themselves Cauchy distributed. In other words, the Cauchy distribution is a stable distribution.
Notice that the central limit theorem doesn't apply to Cauchy distributed random variables because they don't have finite mean and variance.
A: If $X_1, \ldots, X_n$ are i.i.d. Cauchy$(0, 1)$ then we can show that $\bar{X}$ is also Cauchy$(0, 1)$ using a characteristic function argument:
\begin{align}
\varphi_{\bar{X}}(t) &= \text{E} \left (e^{it \bar{X}} \right ) \\
&= \text{E} \left ( \prod_{j=1}^{n} e^{it X_j / n} \right ) \\
&= \prod_{j=1}^{n} \text{E} \left ( e^{it X_j / n} \right ) \\
&= \text{E} \left (e^{it X_1 / n} \right )^n \\
&= e^{- |t|}
\end{align}
which is the characteristic function of the standard Cauchy distribution. The proof for the more general Cauchy$(\mu, \sigma)$ case is basically identical.
