There is this theorem that states:
If $\alpha$ is an ordinal, then there is a limit ordinal $\mu$ with $\alpha < \mu$.
The definition of limit ordinal is just "an ordinal which is a limit", but a better reformulation of this definition is:
Let $\lambda$ be a non-zero ordinal. $\lambda$ is a limit ordinal if and only if for all ordinal $\alpha$, if $\alpha<\lambda$ then $\alpha^+<\lambda$. ($\alpha^+$ is the successor ordinal of $\alpha$)
I don't understand how that theorem makes sense, the book I'm reading gives the proof as an exercise (which I can't do) so I can't even see why it's true. If for example I have the ordinal $1=\{\emptyset\}$ what is a limit ordinal for this? Is it $\omega$ for example (or $2\omega$, $3\omega$ etc)? Is this what it means?
Yes, the ordinals $\omega$, $\omega2$, $\omega3$ etc. work for $\alpha=1$.
In general, for any $\alpha$, the ordinal $\alpha+\omega$ works (or, more generally, the ordinal $\alpha+\beta$, where $\beta>0$ is a limit ordinal).
As a side remark: the ordinals $\omega$, $2\omega$, $3\omega$ etc. are all equal.