0/0 alakú a határérték, ezért tagonként nem megy. Használhatjuk például a L'Hospital-szabályt:

`d/(dn) [sqrt((n+2)/n)-1]``=``d/(dn) [(1+2/n)^(1/2)]``=``1/2*(1+2/n)^(-1/2)*(-2n^-2)``=``-1/(n^2 sqrt(1+2/n))``=``-1/sqrt(n^4+2n^3)`


`d/(dn) [sqrt((4n-1)/n)-2]``=``d/(dn) [(4-1/n)^(1/2)]``=``1/2*(4-1/n)^(-1/2)*n^-2``=``1/(2n^2 sqrt(4-1/n))``=``1/sqrt(16n^4-4n^3)`


`lim_{n rightarrow oo} (sqrt((n+2)/n)-1)/(sqrt((4n-1)/n)-2)``=``lim_{n rightarrow oo} (d/(dn) [sqrt((n+2)/n)-1])/(d/(dn) [sqrt((4n-1)/n)-2])``=``lim_{n rightarrow oo} -sqrt(16n^4-4n^3)/sqrt(n^4+2n^3)``=``lim_{n rightarrow oo} -sqrt((16-4/n)/(1+2/n))``=``-sqrt((16-0)/(1+0))``=-4`