\lim \frac { 2 n } { n + 1 } = 2
Solve for 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
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\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l=2
The equation is in standard form.
\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)}
Divide both sides by 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)}
Dividing by 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}) undoes the multiplication by 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)}
Divide 2 by 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
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