In the epsilon-delta proof for $\lim\limits_{x \to 0} \frac{\sin(x)}{x}$, we can show that for $|x| < 1$, $|\frac{\sin(x)}{x}| < \frac{x^2}{6}$. We can set $\delta = \sqrt{6\epsilon}$ and complete the proof. During the proof, $\delta$ is taken to be the minimum of $\sqrt{6\epsilon}$ and $1$. Why is $\delta$ defined this way, and why is 1 included in the definition of $\delta$? What would go wrong if $\delta$ is just $\sqrt{6 \epsilon}$? Can we use another number other than $1$?
Update:
So now I understand that in the proof, $\delta=\sqrt{6\epsilon}$ is guaranteed to work when $|x|<1$. So we need to deal with the case when $|x| \geq 1$. We want to choose a $\delta$ such that $|x| < \delta$ still yields $$\left|\frac{\sin(x)}{x} – 1\right| = |\frac{x^2}{3!} – \frac{x^4}{5!} + \frac{x^6}{7!} – \frac{x^8}{9!} + \frac{x^{10}}{11!} – \cdots | < \frac{x^2}{6} < \epsilon$$
My question is how do we deal with this case? If there is an upper bound on $\delta$, how do we find it? What is a formal proof?
Also, for $|x| \geq 1$ are we just saying that $\delta = 1$ works or is it actually the case that we are taking $\delta = \min(1, \sqrt{6\epsilon})$? If we set $\delta=1$, note that $|x| < 1$ would be false, so the whole statement
$$0<|x|<\delta \Rightarrow \left|\frac{\sin(x)}{x}-1\right| < \epsilon$$ would be vacuously true.
Best Answer
A common aspect when writing limit proofs is we often tend to write the proof backwards.
That is, we explain how he found a $\delta$ that works.
But that is more a communication style, showing how we found a value.
The formal way to prove the limit need no "how we found it" argument, we just start with:
From there, we note that if $|x-0|<\delta$ then $|x|<1$ and thus we make the argument about alternating decreasing sequences, so we get:
$$1\geq \frac{\sin x}{x}\geq 1-\frac{x^2}6.\tag1$$
So:
$$0\leq 1-\frac{\sin x}{x}\leq \frac{x^2}6.$$
But because $|x|< \delta,$ we also have $|x|<\sqrt{6\epsilon}$ so $\frac{x^2}6<\epsilon.$
So $\delta$ works.
We used $1$ because it is really obvious that $\frac{x^{2n}}{(2n+1)!}$ is decreasing when $|x|<1.$
But we could also have used $\delta=\min(\sqrt 6,\sqrt{6\epsilon}).$ It just would have required more work to prove the alternating sequence rule applies.
In general, in any limit proof, if $\delta>0$ works, then any smaller positive $\delta'$ works too. The $\delta$ you find is always one of an infinite set of possible values that work. To prove a limit, you don't need the "best" $\delta,$ just one of the infinite possible values.
We usually pick a $\delta$ that makes our argument simplest, so we use $\min(1,\dots)$ here when $\min(2,\dots)$ would work, to make our argument easier.
But we can't just use $\delta=\sqrt{6\epsilon}$ because there can be values in $|x|<\delta$ where the alternating sum argument won't work.
I think you could use $\delta=\min(\sqrt{20},\sqrt{6\epsilon})$ and still get a proof, although the series doesn't always decrease in the first two terms for all $x$ in the range. I believe ou can still show $(1)$ is true, which is all you really need, but the proof is a bit more complicated.
So $1$ is chosen just to make our life easier. We could have used $2$ or $\sqrt{20}$ or $1/1000000$ or $1/\pi.$ But $1$ makes our argument simple, so we choose that.