Can I pick this value for $\delta$?

Question

Prove $$\lim_{z\to z_0}(z^2+c)=z_0^2+c$$

I tried $$|z^2+c-(z_0^2+c)|=|z^2-z_0^2|\leq|z|^2+|z_0|^2\leq|z+z_0-z_0|^2+|z_0|^2\leq|z-z_0|^2+2|z_0|^2<\epsilon.$$

Can I pick $\,\delta = \sqrt{\epsilon-2|z_0|^2}\,$?

Answer 1

No, because then $\epsilon-2|z_{0}|^{2}$ may be a negative number.

For a proof, We let $\delta=\min\{1,\epsilon/(1+2|z_{0}|)\}$, then $|z^{2}+c-(z_{0}^{2}+c)|=|z-z_{0}|\cdot|z+z_{0}|\leq(1+2|z_{0}|)|z-z_{0}|<\epsilon$.