lim n-> inf din n(√(n²+5)-√(n²+1))=
=lim n->inf n(√(n²+5)-√(n²+1))*(√(n²+5)+√(n²+1))/(√(n²+5)+√(n²+1))=
=lim n->inf n(n²+5-(n²+1))/(√(n²+5)+√(n²+1))=
=lim n->inf n*4/(n(√(1+5/n²)+n√(1+1/n²))=
=lim n->inf 4n/n(√(1+5/n²)+√(1+1/n²))=
=lim n->inf 4/√(1+5/n²)+√(1+1/n²)=
=4/(1+1)=4/2=2
L=2 => L∈(1;3)
