\lim \frac { 2 n } { n + 1 } = 2
Réitigh do l.
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\neq 0\text{ and }n\neq -1
Roinn
Cóipeáladh go dtí an ghearrthaisce
\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l=2
Tá an chothromóid i bhfoirm chaighdeánach.
\frac{\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
Roinn an dá thaobh faoi 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
Má roinntear é faoi 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}) cuirtear an iolrúchán faoi 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}) ar ceal.
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
Roinn 2 faoi 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
Samplaí
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