# Find the following complex limits. lim_{nrightarrowinfty}frac{6n}{2n-n^2i}

Question
Limits and continuity
Find the following complex limits. $$\lim_{n\rightarrow\infty}\frac{6n}{2n-n^2i}$$

2021-02-03
$$\lim_{n\rightarrow\infty}\frac{6n}{2n-n^2i}=\lim_{n\rightarrow\infty}\frac{6}{2n-ni}$$
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 $$\frac{6}{2-ni}=\frac{12}{n^2}+\frac{\frac{6}{ni}}{\frac{(4^2)}{n^2+1}}$$
$$=\frac{0+0i}{(0+1)}$$
$$=0$$

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