# Use $\varepsilon$-$\delta$ Definition of Limits to Show a Statement is True

I am trying to show that the following statement is true: $$\lim_{z\to z_0}\left(z^2 + c\right) = z_0^2+c$$ where $z$, $z_0$, and $c$ are complex. To show this, I am supposed to rely on the following definition of a limit: $$\lim_{z\to z_0} f(z) = w_0$$ where $w_0$ is also complex, and $f(z)$ is defined in some deleted neighborhood of $z_0$, means that for each positive number, $\varepsilon$, there is a positive number $\delta$, such that $$\lvert f(z) - w_0 \rvert < \varepsilon \qquad\mathrm{whenever}\qquad 0<\lvert z-z_0 \rvert < \delta.$$

My approach is as follows. Assume $\lvert z-z_0 \rvert < \delta$, and $\varepsilon > 0$. \begin{align} \lvert (z^2+c) - (z_0^2 + c) \rvert &= \lvert z^2-z_0^2\rvert\\&=\lvert (z-z_0)(z+z_0) \rvert\\ &=\lvert z-z_0\rvert \lvert z+z_0 \rvert \\ &= \lvert z-z_0 \rvert \lvert z-z_0 + 2z_0 \rvert \\ & \leq\lvert z-z_0\rvert (\lvert z-z_0 \rvert +\lvert 2z_0 \rvert) \qquad \mathrm{by\,triangle\,inequality}\\ & < \delta^2+2\delta\lvert z_0\rvert \end{align} Now, if I define $\varepsilon\equiv \delta^2 + 2\delta \lvert z_0 \rvert$, then I can say that $$\lvert (z^2+c) - (z_0^2+c)\rvert<\varepsilon \qquad \mathrm{whenever}\qquad \lvert z-z_0\rvert < \delta.$$ My question is: Did I do anything wrong? This seems much more involved than the previous exercises I've worked through, and so I'm also wondering if there's a better/easier way to solve this problem.

If you want me to be nitpicky, I'd say that you should clarify your line "I define $\varepsilon \equiv \delta^2+2\delta|z_0|$," since you're really using this for a given, fixed $\varepsilon$ to define $\delta$. If you're just starting with proofs, it might also be worth noting that the $\delta$ so defined is a new entity, distinct from the $\delta$ that appears above. So it might be worth giving this one a different letter unless you're sure that your reader won't mind. Possible choices of rephrase are: "Choose a $\delta_0>0$ such that $\delta_0^2 + 2\delta_0|z_0| < \varepsilon$" (or equal to $\varepsilon$) or "Making $\delta$ smaller, if necessary, we may suppose that $\delta^2 + 2\delta|z_0|< \epsilon$", if you want to overload the $\delta$ used above.
• Thank you very much for your answer. I really want to be able to write proofs that can survive a "nit-picking". So, I must ask more about the nitpicky stuff above. First, I think I understand the part about "you're really using this for a given, fixed $\varepsilon$ to define $\delta$." In other words, you're saying that since $\varepsilon$ is the independent variable in the definition of a limit, then $\delta$ should depend on the choice of $\varepsilon$, and not the other way around. Does that sound right? – Mike Bell May 6 '16 at 18:26
• And for the second part, about "the $\delta$ so defined is a new entity, distinct from the $\delta$ that appears above," I don't totally understand this. If (and maybe this is not the right assumption to make) the $\varepsilon$ in my proof is the same as that in the definition of a limit, then will that not require that the $\delta$ that shows up in my proof also be the same $\delta$ that showed up in the definition of the limit? Thank you! – Mike Bell May 6 '16 at 18:30
• Hey Mike. Yes, your first part sounds right. Your $\delta$ should be a response to an $\varepsilon$. So, given $\varepsilon$, you can define a $\delta$ such that... whatever. I just wanted to make sure that it was clear that your equation is restricting/defining the $\delta$. – Josh Keneda May 7 '16 at 8:30
• The second part isn't really worth worrying about. I was noting that your argument starts with "Assume $|z-z_0|< \delta$," and then you define $\delta$ a few lines later. Either this is a new $\delta$ or it's being defined after already in use. This doesn't actually cause any problems in your argument, but you could consider reorganizing for clarity. You could bring your definition of $\delta$ to the front of your argument, or you could use a different letter when deriving the inequalities. – Josh Keneda May 7 '16 at 8:46