Prove for definition $\lim_{z\to-i}z^2+1=0$

Question

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Proof: By definition

$$\lim_{z\to-i}z^2+1=0$$

if given $\epsilon >0$ I can find $\delta>0 $ such that,.

If $0<|z+i|<\delta$ then $|z^2+1|<\epsilon $

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$$|z^2+1|=|z^2-i^2|=|(z-i)(z+i)|<\delta |z-i|$$

then??

Answer 1

Start by assuming that $|z+i|<1$. Then by the triangle inequality we get

$$\begin{align}|z-i|&=|(z+i)+(-2i)|\\ &\leq |z+i|+|-2i|\\ &=|z+i|+2\\&<1+2\\&=3. \end{align}$$

Choose $\delta=\min\{1,\epsilon/3\}$. Then, $\delta\leq 1$ and $\delta\leq \frac{\epsilon}{3}$. Thus, if $0<|z+i|<\delta$ then $$|z^2+1|=|z^2-i^2|=|(z-i)(z+i)|<\delta |z-i|<3\delta\leq \epsilon.$$ Done.