# Thread: Help with epsilon-delta proof of sin x / x = 1 as x ->0

1. ## Help with epsilon-delta proof of sin x / x = 1 as x ->0

Hello I am getting stuck on how to prove the limit sin x / x = 1 as x ->0 with epsilon delta.

I have abs abs((sin x / x) -1)< epsilon and since sin x / x will always be less then 1 it would just become (1- sin x / x) < epsilon right?

also abs(x - a) < delta would just be abs(x) < delta since a = 0.

The goal now is to get the inequality with epsilon simplified to just x?

2. This is usually proved not by definition, but from geometric arguments. See PlanetMath, here, at Dr. Math Forum with some pretty bad ASCII notation. You can also Google for "sin(x) / x".

I have abs abs((sin x / x) -1)< epsilon and since sin x / x will always be less then 1 it would just become (1- sin x / x) < epsilon right?
Yes, though this does not move you very far.

3. Thanks for the reply. I only tried to tackle the problem b/c of seeing it proven through the squeeze theorem prior. I wanted to prove to myself that all limits that exist can be proven through epsilon delta but the first one that came to mind sin x / x I cant seem to figure out.

4. I think that the squeeze theorem has a constructive proof. Namely, suppose that $\displaystyle g(x) \leq f(x) \leq h(x)$ and $\displaystyle \lim_{x \to a} g(x) = \lim_{x \to a} h(x)$. If one knows how to find $\displaystyle \delta$ from $\displaystyle \epsilon$ in the definitions of limits for g and h, then the proof provides a method for finding $\displaystyle \delta$ from $\displaystyle \epsilon$ in the definitions of limits for f. Thus, one can start with some facts from trigonometry, such as $\displaystyle \sin(x)<x<\tan(x)$, and by inspecting the proof of the squeeze theorem, to find the function $\displaystyle \delta(\epsilon)$ for the limit of $\displaystyle \sin(x)/x$.

5. Hmm... so functions like sin x / x as x -> 0 can only be proven through squeeze? There are no algebraic manipulations that can directly prove that there is always a delta for any given epsilon?

6. I found another article on Dr. Math which has a direct proof. It uses the fact that $\displaystyle \sin(x) > x - x^3/3!$ for x > 0. I am not sure how to prove this without Taylor series (using Taylor series may involve circular reasoning; this has to be checked carefully).

But what I was trying to say is that there is no such thing as "squeeze theorem" in the ultimate picture of things. Every fact in calculus can be proved from axioms of real numbers and definitions. Suppose that instead of invoking the squeeze theorem, you inserted its proof in your argument. Then, when the whole argument is considered, who is to say where the squeeze theorem begins and ends? Besides, when one considers a specific example, one can apply simplifications or avoid some cases that have to be considered in a general proof. So what you are looking at may have the origin in the squeeze theorem, but it may be simplified, compressed, rearranged, etc. It may be pretty similar to the argument in Dr. Math above.

What is important is the final argument tells you how to build a delta for each epsilon specifically for sin(x)/x.

I am not sure if this philosophical rumbling is useful...

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### sin(x)/x epsilon delta

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