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this is my first question and it is about an exercise of calculus 1 I have seen at the University when I was an undergraduate. Till today I have not found the answer.

The exercise is the following: calculate the derivative of $\sin(x)/x$ for $x \to 0$ without using neither L'Hôpital's rule nor Taylor expansion.

This leads to: $$\lim_{x\to0}\dfrac{\sin(x)/x-1}{x}=\lim_{x\to0}\dfrac{\sin(x)-x}{x^2}$$

Using l'Hopital rule or Taylor expansion (or even drawing the function) is easy to see the result is 0 but I would be curious to see how to get it otherwise. I have tried to use the squeeze theorem but without success so far.

Many thanks in advance to those who will provide insight here :-)

It's a feature and not a bug
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  • A well-known proof is using geometry and squeeze theorem. See, for ex., https://www.maths.tcd.ie/~lebed/Remarkable.pdf – Vasili May 02 '23 at 16:38

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You could rewrite the limit to resemble the derivative in 0:

$$ \lim_{x\to0} \frac{\sin(x)}{x} = \lim_{x\to0} \frac{\sin(0 + x) - \sin(0)}{x} = \sin'(0) = \cos(0) = 1. $$

omicron
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    How do you prove that the derivative of $\sin$ is $\cos$ ? Usually it's done using the limit you are trying to prove so it would be a circular reasoning. – Lelouch May 02 '23 at 17:07