Just asking for verification of this proof. I've tried to be as clear as possible about exactly what I'm doing, but I'm struggling in this subject so let me know if I've done anything incorrectly or made any assumptions that require more justification.
I understand why $\delta = \min \{2, \dfrac{\epsilon}{28}\}$, but I'm a little shaky on reasoning what is happening when $\delta = 2$. Most of the proofs I've seen that are similar to this don't really elaborate on the case when $\delta$ is that large. Any insight on that is also very appreciated.
Prove that $f(x) = x^3 + 1$ is continuous at $a = 2$.
Choose $\epsilon > 0$ and let $|x-2| < \delta \leq 2$. Then $$|x-2| < 2 \Leftrightarrow -2 < x - 2 < 2 \Leftrightarrow 0 < x < 4$$ Let $\delta = \min \{2, \dfrac{\epsilon}{28}\}$.
If $\dfrac{\epsilon}{28} < 2$ Then we have $$|x - 2| < \delta$$ $$|x - 2| < \dfrac{\epsilon}{28}$$ $$|x - 2| \cdot 28 < \epsilon$$ $$|x - 2| |x^2 + 2x + 4| < \epsilon$$ $$|x^3 - 8| < \epsilon$$ $$|x^3 + 1 - 9| < \epsilon$$ $$|f(x) - f(2)| < \epsilon$$
If $2 > \dfrac{\epsilon}{28}$, then $\delta = 2$. Then $\epsilon > 64 = |f(4) - f(0)|$, which is true when $|x-2| < \delta = 2$.
Thus, $f(x) = x^3 + 1$ is continuous at $a = 2$.
For continuity questions, given $\epsilon$, one needs to find an appropriate $\delta$. The choice of $\delta \le$ 2 is arbitrary; in the sense that if you used any other value, and repeated the same calculations, you would find an upper bound for $\epsilon$. The key is to choose a number and then show that taking $\delta$ sufficiently small will work. Many texts, use 1 as an initial value. If you had done that, the bound would have been $\frac{\epsilon}{19}$.
Also in your first line it should be -2 $< \mid x - 2 \mid <$ 2.