Since $\alpha + \beta $ is a limit ordinal, we have that forall $x\in \alpha + \beta \;$, $x+1\in \alpha + \beta \;$
Lets assume that $\beta\;$ is not a limit ordinal, then there exists some $x\in \beta \;$ such that $x+1 > \beta $
Although, if $x\in \beta \;$, then $x\in \alpha + \beta \;$ but $\;\alpha + \beta \;$ is a limit ordinal, then we get that $x+1\in \alpha + \beta \;$
Since $x+1 > \beta \;$ we get that $\alpha+\beta<\alpha+x+1$.
That's how far I have managed to reach in my general direction of the proof, I wasn't able to conclude any contradiction, altough intuitivly I totally understand why the statement is true.
I would love to get some help!
Hint: If $\beta$ is no limit ordinal, it has a maximum, and then $\alpha+\beta$ has a maximum as well since $\alpha+\beta$ is just $\alpha$ with $\beta$ concatenated after it.