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 $1$?


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.

  • $\begingroup$ Because when $|x|>1,$ you don't know the coefficients are decreasing in absolute value, so you don't know that the value of the series is greater than the 2nd partial sum. $\endgroup$ Jan 13 at 20:15
  • $\begingroup$ is the choice of $1$ arbitrary or it is because $1$ is the radius of convergence? Also based on the argument, $\delta < 1$ and $\delta < \sqrt{6\epsilon}$. The latter is obvious. Why do we have former one too: $\delta < 1$? $\endgroup$
    – Josh
    Jan 13 at 20:18
  • $\begingroup$ In particular, if $x^6/7!>x^4/120$ then it is possible for $1-\frac{\sin x}x>x^2/6.$ So you really need $x^2\leq 42,$ I think. So $1$ isn't necessary, we could use $\sqrt{42}.$ $\endgroup$ Jan 13 at 20:20
  • $\begingroup$ The radius of convergence is $+\infty$, but one would need to prove that. See math.stackexchange.com/questions/577676/… $\endgroup$
    – Andrei
    Jan 13 at 20:20
  • $\begingroup$ Well, $\delta\leq 1$ is all we really need. $\endgroup$ Jan 13 at 20:20

2 Answers 2


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:

Given $\epsilon>0$ let $\delta=\min(\sqrt{6\epsilon},1).$

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:

$$1\geq \frac{\sin x}{x}\geq 1-\frac{x^2}6.\tag1$$


$$0\leq 1-\frac{\sin x}{x}\leq \frac{x^2}6.$$

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.


For $|x| \leq 1$, we argue that $\left| \frac{\sin(x)}{x} -1 \right| < \frac{x^2}{6}$. So if $\epsilon > \frac{x^2}{6}$ we can conlcude that $|x| < \sqrt{6\epsilon}$. Note that it is really true for $|x|=1$, but the endpoints need to be checked separately. So this argument only works for when $|x| \leq 1$. What about $|x| > 1$? We don't know yet.

To prove $\left| \frac{\sin(x)}{x} -1 \right| < \frac{x^2}{6}$, an argument was used that involved converging alternating series. That is $$\left| \frac{\sin(x)}{x} -1 \right| = \left| \frac{x^2}{3!} - \frac{x^4}{5!} + \frac{x^6}{7!} - \cdots \right| < \frac{x^2}{6}$$ But when $|x| > 1$, we cannot be certain this statement is true so we have to find an alternative (interestingly, I am almost certain that $\left| \frac{\sin(x)}{x} -1 \right| < \frac{x^2}{6}$ is true for all non-zero $x$ but we did not prove this so we cannot use this).

So let $|x|>1$. Let $\delta = 1$. Then $|x| < 1 = \delta$ is false, since $|x| > 1$. Therefore the statement $$0<x<\delta \Rightarrow |f(x)-L| < \epsilon$$ is vacuously true.

So from $|x| \leq 1$, we have $\delta = \sqrt{6\epsilon}$ and from $|x| > 1$ we have $\delta < 1$. Let $\delta=\min(\sqrt{6\epsilon}, 1)$, then we can be certain that

$|x|<1$, which in fact implies that $\left| \frac{\sin(x)}{x} - L \right| < \frac{x^2}{6} < \frac{\left(\sqrt{6\epsilon}\right)^2}{6} = \epsilon$, and we are done.

The takeaway message is that, if we look at this argument abstractly, it goes something like:

Let $a \in \mathbb R$ and $a > 0$. Let $x \leq a$. Assume that it can be shown that $|f(x) - L| < g(x)$, where $g(x)$ is some function that has some desirable properties in the context of the specific problem that is being solved. Let $\epsilon > g(x)$. Again assume that by some simplification, it can be shown that $|x| < \delta(\epsilon)$.

Then for $|x| > a$, let $\delta = a$. There is no $x$ that satisfies the inequality $|x| < a$, so it follows that $$|x| < \delta \Rightarrow |f(x) - L| < \epsilon$$ The specific values of $f(x)$ and $L$ are not important in this case.

Let $\delta = \min(\delta(\epsilon), a)$. Then $$|x| < \delta \leq \delta(\epsilon) \Rightarrow |f(x) - L| < \epsilon$$

The fact that $1$ is the obvious choice in this example is because it was shown for $|x| \leq 1$, we can find $\delta$ such that $|f(x)-L|<\epsilon$. This was the most challenging part of the proof, so the fact that for $|x|>1$, we can just take $\delta=1$ and the implication being vacuously true may sound a bit like cheating, but it is the reward of the hardwork of the first part.

A more challenging question to ask would be:

Can we pick a larger value for $\delta$ when $|x|>1$? In this case we know the answer is yes. And If so, how large can it be? The answer to these questions are not obvious in general. Actually, the answer to the latter question for this specific example is not so obvious unlike the former one (I would appreciate if someone can provide a full proof of this).

But the definition of $\epsilon$-$\delta$ is not concerned with the optimal values of $\delta$. We just have to find a $\delta$ that works.


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