Prove $\displaystyle{\lim_{x \to 2}}\,x^3 + 1 = 9$ $(\delta < 1 \text{ or } \delta \leq 1)?$ My math prof demonstrated this proof today and I'm not sure I understand all the steps. I don't understand why the $\delta$ has to be divided by 2 after selecting the minimum. Here are the steps he showed:
We should show $\forall\epsilon > 0, \exists\delta > 0$ (let's call this eq1):
$$|x-2| < \delta \implies |x^3 + 1 - 9| = |x-2||x^2 + 2x + 4| < \epsilon$$
Assume $\delta < 1$, this bounds $|x - 2| < 1$ and implies $1 < x < 3$. Now we know:
$$|x^2 + 2x + 4| < 19$$
So, if we can show:
$$|x-2| < \delta \implies |x-2||x^2 + 2x + 4| < 19|x-2| < \epsilon$$
eq1 follows for $\delta < 1$. The above is trivially true for $\delta = \epsilon/19$. But we must account for $\delta >= 1$ and therefore set:
$$
\delta = \frac{1}{2}\min(\epsilon/19,1)
$$
QED. Why is the $\frac{1}{2}$ factor needed? To me it looks redundant. According to the prof it was because $\delta = 1$ is a possibility. But if $\delta = 1$ then $\epsilon = 19$ and:
$$|x-2| < 1 \implies |x-2||x^2 + 2x + 4| < 19$$
is obviously true. Could the $\frac{1}{2}$ factor have been avoided by instead assuming $\delta <= 1$?
I'm not even sure why I have to "soil" my proof by covering the $\delta >= 1$ case with the $\min$ function. Can't I just say "The proof holds for small deltas ($\delta < 1$) which is all that matters for limits?"
 A: You are perfectly right. There is no need to have that factor of $1/2$.  This is not a personal comment on your professor, but in this case his argument is clearly wrong. We don't actually assume $\delta<1$, but rather assume that $|x-2|<1$. Based on this constraint on the values of $x$ we can see that $|f(x) - L|<19|x-2|$. If we further constrain values of $x$ by $|x-2|<\epsilon/19$ (in addition to earlier constraint $|x-2|<1$) then we have the desired inequality $|f(x) - L|<\epsilon$. The constraints on $x$ can be combined as $|x-2|<\min(1,\epsilon /19)$. And hence if we take $\delta=\min(1,\epsilon /19)$ then we know that the following implication holds $$0<|x-2|<\delta\Rightarrow |f(x) - L|<\epsilon$$ In the above we have taken $f(x) =x^{3}+1,L=9$. Also note that if one $\delta$ works any smaller $\delta$ also works so the answer $\min(1,\epsilon/19)/2$ is also correct, but this is not a necessity (as your professor wants to argue about the case $\delta\geq 1$).
Remember that we don't assume anything about $\delta$ (apart from it being positive) and rather evaluate it in terms of $\epsilon$ based on a careful analysis of inequalities involved. Also note that there is no unique expression for $\delta$ in such problems and answers can vary. 
A: When we use $\epsilon-\delta$ language to prove thing, what really matter is the existence of $\delta$. So assume that you could find out some feasible candidates of $\delta$ you can always choose any other $\delta$ which less than them. 
