\(\lim_{n\rightarrow\infty}\frac{6n}{2n-n^2i}=\lim_{n\rightarrow\infty}\frac{6}{2n-ni}\)

Rationalize \(\frac{6}{2-ni}\) as

\(\frac{6}{2-ni}=\frac{6}{2-ni}\cdot\frac{2+ni}{2+ni}\)

\(=\frac{12+6ni}{4-n^2i^2}\)

\(=\frac{12+6ni}{4+n^2}\)

\(=\frac{\frac{12+6ni}{n^2}}{\frac{4+n^2}{n^2}}\)

\(=\frac{\frac{12}{n^2}+\frac{6}{ni}}{\frac{4^2}{n^2}+1}\)

Substitute \(\frac{6}{2-ni}=\frac{12}{n^2}+\frac{\frac{6}{ni}}{\frac{(4^2)}{n^2+1}}\)

\(=\frac{0+0i}{(0+1)}\)

\(=0\)

Rationalize \(\frac{6}{2-ni}\) as

\(\frac{6}{2-ni}=\frac{6}{2-ni}\cdot\frac{2+ni}{2+ni}\)

\(=\frac{12+6ni}{4-n^2i^2}\)

\(=\frac{12+6ni}{4+n^2}\)

\(=\frac{\frac{12+6ni}{n^2}}{\frac{4+n^2}{n^2}}\)

\(=\frac{\frac{12}{n^2}+\frac{6}{ni}}{\frac{4^2}{n^2}+1}\)

Substitute \(\frac{6}{2-ni}=\frac{12}{n^2}+\frac{\frac{6}{ni}}{\frac{(4^2)}{n^2+1}}\)

\(=\frac{0+0i}{(0+1)}\)

\(=0\)