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

Find the following complex limits. $\underset{n\to \mathrm{\infty }}{lim}\frac{6n}{2n-{n}^{2}i}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jaylen Fountain
$\underset{n\to \mathrm{\infty }}{lim}\frac{6n}{2n-{n}^{2}i}=\underset{n\to \mathrm{\infty }}{lim}\frac{6}{2n-ni}$
Rationalize $\frac{6}{2-ni}$ as
$\frac{6}{2-ni}=\frac{6}{2-ni}\cdot \frac{2+ni}{2+ni}$
$=\frac{12+6ni}{4-{n}^{2}{i}^{2}}$
$=\frac{12+6ni}{4+{n}^{2}}$
$=\frac{\frac{12+6ni}{{n}^{2}}}{\frac{4+{n}^{2}}{{n}^{2}}}$
$=\frac{\frac{12}{{n}^{2}}+\frac{6}{ni}}{\frac{{4}^{2}}{{n}^{2}}+1}$
Substitute $\frac{6}{2-ni}=\frac{12}{{n}^{2}}+\frac{\frac{6}{ni}}{\frac{\left({4}^{2}\right)}{{n}^{2}+1}}$
$=\frac{0+0i}{\left(0+1\right)}$
$=0$