Using $\epsilon$-$\delta$ definition, show that $\lim_{x \to 2} x^{2} = 4$.
The proof according to Example 7: Evaluating a limit using the definition
$|x^{2}-4|<\epsilon \implies |x-2|<\frac{\epsilon}{|x+2|}$
Let $\delta < 1$, then $|x-2|<1$
$-1<x-2<1$
$1<x<3$
$3<x+2<5$
$\frac{1}{5}<\frac{1}{|x+2|}<\frac{1}{3}$
$\frac{\epsilon}{5}<\frac{\epsilon}{|x+2|}<\frac{\epsilon}{3}$
Let $\delta \leq \frac{\epsilon}{5}$, then $|x-2| < \frac{\epsilon}{5}$
$|x^{2}-4| < |x+2|\frac{\epsilon}{5}$
As long as $\delta < 1$, we get:
$|x^{2}-4| < |x+2|\frac{\epsilon}{5}< |x+2|\frac{\epsilon}{|x+2|} $
$|x^{2}-4| < \epsilon$
However, we could have let $\delta \leq 1$, which will make the proof still true. Consider the following two cases:
Case I: $\delta < 1$
$\delta < 1 \implies |x-2|< \delta < 1 \implies |x-2|<1$.
Case II: $\delta \leq 1$
$\delta \leq 1 \implies |x-2|< \delta \leq 1 \implies |x-2|<1$ for either $\delta < 1$ and $\delta = 1$
Thus, $\delta \leq \frac{\epsilon}{5} \leq 1 $
Furthermore, the source states:
For the final fix, we set $\delta$ to be the minimum of $1$ and $\frac{\epsilon}{5}$. This way all caculations above work.
This would mean $\delta \leq min(1, \frac{\epsilon}{5})$ since, by definition, a smaller $\delta$ such as $0<|x-a|<\delta^{'}<\delta$ will imply that $|f(x)-L| < \epsilon$ and that $1$ or $\frac{\epsilon}{5}$ will satisfy the implication depending on $\epsilon$. Therefore, we can have $\delta \leq 1$
Question
In summary, is it better to start off with $\delta \leq 1$ over $\delta < 1$, right?