Is this limit proof right?
Solution 1:
You have the right idea. Why did you take $n(\epsilon)=\frac\epsilon{|\alpha|}$, and where was that used? It should be the same $n(\epsilon)$ that was used in the first limit. It should be more along the lines of:
For all $\epsilon>0$, there exists $n(\epsilon)>0$ such that if $n>n(\epsilon)$, then $|a_n-a|<\frac\epsilon{|\alpha|}$, and therefore,
$$|\alpha a_n-\alpha a|=|\alpha||a_n-a|<|\alpha|\frac\epsilon{|\alpha|}=\epsilon.$$