>>12173850>So, the limit of must be 0This is not what the proof implies. If is the zero sequence, then the limit in the assumption doesn't approach 0. Furthermore, one can verify pretty easily that it isn't possible for to approach 0 and a be nonzero nor to approach a nonzero real but a approach zero, and must converge, otherwise the limit ends up being 1 as outgrows a to even matter.
In this way, the proof can be done using some theorems about the algebra of limits (ie, the above implies must approach something nonzero and a != 0 in order to satisfy the first condition) to say then the limit in the denominator is exists and is nonzero, and similarly the limit of the numerator exists, and we can suppose that for some . Then we can take the limit to simply be
and conclude the proof, but again, this requires some results about limits of sequences
