This is exercise from some book:
I have to prove by definition with $\epsilon $ and $ \delta$ the following limit:
$\lim_{z\to 1+i} \frac{1}{z-i} = 1$
What i've tried so far is that:
Let $\epsilon > 0$ we have to find $\delta > 0$ such that $0<|z-(i+1)|<\delta$
such that
$|\frac{1}{z-i}-1| <\epsilon$
$|\frac{z-(1+i)}{z-i}| <\epsilon$
$|\frac{1}{z-i}||z-(1+i)| <\epsilon$
And from here I've got stuck.
Can anyone help me with that?
Best Answer
Suppose that $|z-(1+i)|<\frac12$. Then$$|z-i|=|z-(1+i)+i|\geqslant\bigl||i|-|z-(1+i)|\bigr|>\frac12.$$But then$$\frac{|z-(1+i)|}{|z-i|}\leqslant2|z-(1+i)|.$$So, take $\delta=\min\left\{\frac12,\frac\varepsilon2\right\}$.