Expert answers to "Find the following complex limits. limn→∞13+ni"

texelaare

Answered question

2021-02-03

Find the following complex limits.

lim_{n \to \infty} \frac{1}{3 + n i}

Viktor Wiley

Skilled2021-02-04Added 84 answers

lim_{n \to \infty} \frac{1}{3 + n i}

Rationalize

\frac{1}{3 + n i}

\frac{1}{3 + n i} = \frac{1}{3 + n i} \cdot \frac{3 - n i}{3 - n i}

= \frac{3 - n i}{3^{2} - n^{2} i^{2}}

Substitute

i^{2} = - 1

and solve further

\frac{1}{3 + n i} = \frac{3 - n i}{3^{2} + n^{2}}

= \frac{\frac{3 - n i}{n^{2}}}{\frac{3^{2} + n^{2}}{n^{2}}}

= \frac{\frac{3}{n^{2}} - \frac{1}{n i}}{\frac{3^{2}}{n^{2}} + 1}

Substitute

\frac{1}{3 + n i} = \frac{\frac{3}{n^{2}} - \frac{1}{n} i}{\frac{3^{2}}{n^{2}} + 1}

Hence

lim_{n \to \infty} \frac{1}{3 + n i} = lim_{n \to \infty} \frac{\frac{3}{n^{2}} - \frac{1}{n i}}{\frac{3^{2}}{n^{2}} + 1}

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

= 0

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