Use $\epsilon$ - $\delta$ definition to prove $\lim_{z\to 1}$ $\frac{z+2}{z+3i}$ $=$ $\frac{3}{1+3i}$

Question

So i need need to prove that $$\lim_{z\to 1} \frac{z+2}{z+3i} = \frac{3}{1+3i}$$ So far my understanding is that we want to calculate $\left|{f(z)-z}\right|$ and manipulate it in such a way that we get $\left|{z-1}\right|$ to appear. I have $$\left|\frac{z+2}{z+3i}-\frac{3}{1+3i}\right|$$ but no matter what way i manipulate or simplify it i can't seem to get it to work out. What should it like when it is simplified to the necessary form?

Answer 1

Note

\begin{align}\left|\frac{z + 2}{z + 3i} - \frac{3}{1 + 3i}\right| &= \left|\frac{(z + 2)(1 + 3i) - 3(z + 3i)}{(z + 3i)(1 + 3i)}\right|\\ &= \left|\frac{(-2 + 3i)z + (2-3i)}{(z + 3i)(1 + 3i)}\right|\\ &= \left|\frac{-2 + 3i}{1 + 3i}\frac{z-1}{z+3i}\right|\\ &= \sqrt{\frac{13}{10}}\frac{|z - 1|}{|z + 3i|}. \end{align}

Now if $|z - 1| < (1/2)\sqrt{10}$, then $$|z + 3i| = |(z - 1) + (1 + 3i)|\ge |1 + 3i| - |z - 1| > \sqrt{10} - \frac{1}{2}\sqrt{10} = \frac{1}{2}\sqrt{10},$$

Thus

$$\sqrt{\frac{13}{10}}\frac{|z - 1|}{|z + 3i|} < \frac{\sqrt{13}}{\sqrt{10}}\frac{2}{\sqrt{10}}|z - 1| = \frac{\sqrt{13}}{5}|z - 1|.$$

We want the last expression to be made less than $\epsilon$ in our $\delta-\epsilon$ argument. This suggests to do the following. Let $\epsilon > 0$ and set $\delta = \min\{(1/2)\sqrt{10}, 5\epsilon/\sqrt{13}\}$. For all $z$, $0 < |z - 1| < \delta$ implies

$$\left|\frac{z + 2}{z + 3i} - \frac{3}{1 + 3i}\right| = \sqrt{\frac{13}{10}}\frac{|z - 1|}{|z + 3i|} < \frac{\sqrt{13}}{5}|z - 1| < \frac{\sqrt{13}}{5}\frac{5\epsilon}{\sqrt{13}}= \epsilon.$$