PF: If given $ \epsilon >0 $, let $ N=1$ so whenever $ n>N $ we have $ |c_n-c|=|c-c|=0 < \epsilon $. Therefore the limit of $ c_n=c $ as $ n \rightarrow \infty $ as required.
This is the answer, but I don't understand why we let $ N=1 $. What would be your solution? Thank you!