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I am trying to prove $\lim_{x\to 2} g(x) = 4$ where $g(x) = x^2$. I found this explanation in Abbott's Understanding Analysis:

Let's show

$$\lim _{x \to 2} g(x)=4$$

where $g(x)=x^{2}$. Given an arbitrary $\varepsilon>0$, our goal this time is to make $|g(x)-4|<\varepsilon$ by restricting $|x-2|$ to be smaller than some carefully chosen $\delta .$ As in the previous problem, a little algebra reveals $$ |g(x)-4|=\left|x^{2}-4\right|=|x+2||x-2|$$ We can make $|x-2|$ as small as we like, but we need an upper bound on $|x+2|$ in order to know how small to choose $\delta .$ The presence of the variable $x$ causes some initial confusion, but keep in mind that we are discussing the limit as $x$ approaches $2 .$ If we agree that our $\delta$ -neighborhood around $c=2$ must have radius no bigger than $\delta=1,$ then we get the upper bound $|x+2| \leqslant|3+2|=5\;\forall x \in V_{\delta}(c)$ Now, choose $\delta=\min \{1, \varepsilon / 5\} .$ If $0<|x-2|<\delta,$ then it follows that $$ \left|x^{2}-4\right|=|x+2||x-2|<(5) \frac{\varepsilon}{5}=\varepsilon$$ and the limit is proved.

I don't understand the point of choosing $\delta = \min\{1, \varepsilon/5\}$. Why is it not simply $\delta = \varepsilon/5$? Also, would it be valid to take $\delta = \min\{2, \varepsilon/6\}$ instead of $\delta = \min\{1, \varepsilon/5\}$? Why/why not?

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Ricky_Nelson
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    Forcing delta to be less than or equal to 1 forces $|x + 2|$ to be no more than 5. This is crucial in showing the expression is less than epsilon. If we did not force delta to be less than 1, we would have no control over $|x + 2|$. Note that if $\delta \leq 1$, then $|x-2| < 1$. This implies $-1 < x-2 < 1$. Adding 4 to all sides of the inequality gives $3 < x+2 < 5$ which implies the estimate $|x+2| < 5$. – Nicholas Roberts May 06 '20 at 17:06
  • I feel like this question is the same as yours with different numbers: https://math.stackexchange.com/questions/3658590/understanding-epsilon-delta-proof – teddy May 06 '20 at 17:08
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    Did you mean $\lim\limits_{x\to\color{red}2}$ (in the question and title)? – J. W. Tanner May 06 '20 at 17:08
  • @J.W.Tanner Yes, I did. I've edited my question. Thanks! – Ricky_Nelson May 06 '20 at 17:09
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    @teddy You're not wrong, that link actually resolved my doubts! – Ricky_Nelson May 06 '20 at 17:21
  • Usually when there's a "min" in the formula for $\delta$ it's to deal with large $\epsilon$s. We're usually worried about what happens for small $\epsilon$s so it may seem unnecessary, but we are supposed to find a $\delta$ for any positive $\epsilon$, large or small.
    • – JonathanZ May 06 '20 at 20:58
  • Values (and formulas) for $\delta$ are very much not unique. (For example, if $\delta_0$ works, then so would $\delta_0/2$, or $\delta_0/100$.) I haven't verified whether your particular formula with $2$ and $6$ is valid, but it easily could be. You'll start to see some common patterns -- like breaking the domain into two parts based on whether the distance is more or less than 1, as opposed more or less than 2 -- and so graders and other readers of your math will find it easier if you use those patterns too. But other choices are valid.
    • – JonathanZ May 06 '20 at 21:02
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    @JonathanZsupportsMonicaC Your comments are quite insightful, thank you! – Ricky_Nelson May 06 '20 at 21:22