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I've been looking over some old calculus stuff, and I came back across the following limit.

When deriving $\frac{d}{dx}\sin(x)$, the identity $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ is used as an intermediate step. I'm aware of the geometric proof, that $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ using the squeeze theorem, but I was wondering if there exists any other noncircular proof of this limit that does not require differentiation.

Of course you can verify this identity with L'Hospital's rule, but that proof would not be valid if the derivative of $\sin(x)$ had yet to be proven.

Further you could modify the Taylor expansion of $\sin(x)$ and take the limit of each term and arrive at the same result, but likewise, it is only possible to know the Taylor expansion when you know the derivative of $\sin(x)$.

My last idea was to compare the left-hand and right-hand limits of the. Indeed this works and provides the result we want, but calculating something on the order of $\sin(0.0001)$ requires a calculator, and calculators calculate such values by Taylor expansion... so again this is unsatisfactory to me.

I say all this to pose this question. Aside from the standard geometric proof using the squeeze theorem and the unit circle, is there any other way I can prove this limit with non-circular logic?

MattKuehr
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    Whats the definition of sine you want to use? – stochs May 20 '24 at 04:13
  • @stochs forgive me if I misunderstand your question. If you're asking if I am choosing $\sin(x)$ to be a function of radians or degrees, then radians. – MattKuehr May 20 '24 at 04:15
  • I had a similar question a while ago. It comes down to the question: what does $\sin(x)$ mean? https://math.stackexchange.com/questions/4407632/can-we-prove-lim-x-to0-frac-sin-xx-1-with-a-functional-definition-for – hellofriends May 20 '24 at 04:18
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    I believe @stochs meant, are you defining sin(x) by its Taylor series? By its geometric properties? Or some other way? – H. sapiens rex May 20 '24 at 04:19
  • In that case, by the geometric properties. Thanks for the clarification @H.sapiensrex – MattKuehr May 20 '24 at 04:25
  • @hellofriends That link is very helpful, thank you. Not sure how I couldn't come across it when I initially searched this question... – MattKuehr May 20 '24 at 04:27

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This depends on how you define the sine.

If you define the sine as "opposite over hypotenuse", then you are likely bound to the geometric squeeze proof.

A more analytic path is to define the sine via the series and deduce everything else from there. For a bit of self-promotion, here is an answer detailing how one can start from the series and deduce that what the series defines is actually "opposite over hypotenuse".

Martin Argerami
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