Of course you are right, I don't mean what I say at all do I .
What I do mean is that for every epsilon>0, there exists an N such that
for all n>N:
-epsilon< (a_n)/n-L <epsilon
so:
nL-n epsilon< a_n < nL + n epsilon,
or:
n(L-epsilon) < a_n < n(L+epsilon)
so if we choose a small enough epsilon, as n-> infnty a_n is trapped between
two sequences both of which go to infinity.
RonL