You have the right idea, but it isn't written well. You need to use your observations to write your argument in reverse. Something like this:
Given can [itex]\rightarrow[/itex] L we will show that an [itex]\rightarrow[/itex] L/|c|, which is a contradiction.
Suppose[itex]\ \epsilon > 0[/itex]. Then there is N > 0 such that:
| can - L| < |c|[itex]\epsilon [/itex]
for all n > N. Therefore
|can - L| = |c(an - L/c| = |c||(an - L/c| < |c|[itex]\epsilon[/itex]
which gives, upon dividing that last inequality by |c|, | an - L/c| < [itex]\epsilon[/itex]
for all n > N.