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I am asked to solve following limit problem:

$\lim_{x \to 0}\frac{\sin x}{x}$

It is pretty straightforward using L'Hôpital's rule:

Since $$ \lim_{x \to 0}\frac{\sin x}{x} = \frac{0}{0}$$ Then we take derivative of the numerator and denominator, getting: $$ \lim_{x \to 0}\frac{\cos x}{1} =1$$

However, I would like to know whether there is a way to solve the limit problem above without L'Hôpital's rule. Is there?

iwbtrs
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  • Do you know that $\sin \, x$ has a Taylor series expansion around $0$? – Kavi Rama Murthy Aug 12 '19 at 09:56
  • @KaviRamaMurthy, unfortunately, no. – iwbtrs Aug 12 '19 at 09:56
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    Yes, that can be proved without L'Hopital. In fact you'd better, since that limit is usually what you need to find the derivative of $\sin$ in the first place. So using L'Hopital is circular resoning. – Ethan Bolker Aug 12 '19 at 09:57
  • @Nelver: A Google search on "limit sinx/x" yields a number of links to proofs. – quasi Aug 12 '19 at 10:00
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    @Nelver: A search of Math.SE for "[proof limit sin x/x](https://math.stackexchange.com/search?q=proof+limit+sin+x%2Fx)" yields almost a thousand results. (Not all are specifically about this limit, of course.) I suspect you can find something useful among them. – Blue Aug 12 '19 at 10:03

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You could find the proof in any calculus textbook. Or, here, for instance.

https://en.wikipedia.org/wiki/Squeeze_theorem#Second_example