# Find the following complex limits. lim_{nrightarrowinfty}frac{1}{3+ni}

Find the following complex limits. $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3+ni}$
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

Viktor Wiley
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3+ni}$
Rationalize $\frac{1}{3+ni}$ as
$\frac{1}{3+ni}=\frac{1}{3+ni}\cdot \frac{3-ni}{3-ni}$
$=\frac{3-ni}{{3}^{2}-{n}^{2}{i}^{2}}$
Substitute ${i}^{2}=-1$ and solve further
$\frac{1}{3+ni}=\frac{3-ni}{{3}^{2}+{n}^{2}}$
$=\frac{\frac{3-ni}{{n}^{2}}}{\frac{{3}^{2}+{n}^{2}}{{n}^{2}}}$
$=\frac{\frac{3}{{n}^{2}}-\frac{1}{ni}}{\frac{{3}^{2}}{{n}^{2}}+1}$
Substitute
$\frac{1}{3+ni}=\frac{\frac{3}{{n}^{2}}-\frac{1}{n}i}{\frac{{3}^{2}}{{n}^{2}}+1}$
Hence
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3+ni}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{3}{{n}^{2}}-\frac{1}{ni}}{\frac{{3}^{2}}{{n}^{2}}+1}$
$=\frac{\left(0-0i\right)}{\left(0+1\right)}$
$=0$