I would love some advice on how to approach the following limit:
For each complex number z, determine:
$$ \lim_{n \to \infty}\frac{1}{1+n^{2}z} $$
What i tried is the following:
Let's note the following
$$\left| \frac{1}{1+n^{2}z} \right| = \frac{1}{\left| 1 + n^{2}z\right|} = \frac{1}{\left| 1 + n^{2}(r(cos(\theta)+i(sen(\theta)))\right|} = \frac{1}{\sqrt{(1+n^{2})r^{2}cos(\theta)^{2} +r^{2}sen(\theta)^{2}}} \lt 1$$
$$\left| z_n \right|\to 0$$
therefore $$ z_n \to 0$$
then the limit is equal to 0, is my idea correct?
You can avoid applying the polar coordinates. If $z\neq 0$ by the triangle inequality we have $$|1+n^2z|\ge n^2|z|-1>0,\ {\rm for}\ n>|z|^{-1/2}$$ Hence $$|x_n|\le {1\over n^2|z|-1}\to 0$$