= 3*lim cand x tinde la -∞ din x+ (1/2) *lim x d tinde la -∞ din sin2x= 3*(-∞) +un sir fara limita, dar marginit intre [-1 si1] ;
se putea obtine si din sinx si cos x∈[-1;1] deci si produsul lor ∈[-1;1]
deci limita ceruta este
=-infinit +ceva variabil dar ∈[-1;1]=-infinit
pt ca
daca la -infinit insumez algebric diverse numere finite, cuprinse intr-un interval marginit, suma algebrica tot catre - infinit se duce
exemple numerice
-10000000000 +1 , -1000000000000000000000000-1; -300000000000000000000000000+1/2
-4000000000000000000000000000000000000-1/2
-6000000000000000000000000000000000000000000+√3/2
-9000000000000000000000000000000000000000000000000-√2/2
ai prins ideea , sper ,nu mai conteaza ce adaug/ scad ceva finit la un nr ffffffffffffffffffffffffffffffffffffffffffff mare sinegativ, nu mai influenteaza cu nimic
