\lim \frac { 2 n } { n + 1 } = 2
Whakaoti mō 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
Tohaina
Kua tāruatia ki te papatopenga
\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l=2
He hanga arowhānui tō te whārite.
\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)}
Whakawehea ngā taha e rua ki te 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)}
Mā te whakawehe ki te 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}) ka wetekia te whakareanga ki te 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
Whakawehe 2 ki te 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
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