proposition: Lemma is_lim_seq_div' (u v : nat -> R) (l1 l2 : R) : is_lim_seq u l1 -> is_lim_seq v l2 -> l2 <> 0 -> is_lim_seq (fun n => u n / v n) (l1 / l2).
proof: Proof. intros. eapply is_lim_seq_div ; try eassumption. now contradict H1 ; case: H1 => ->. by []. Qed.
imports: Coq(Reals Psatz ssreflect) Rcomplements Rbar Lub Markov Hierarchy