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I was considering using L'Hopital for $\displaystyle\lim\limits_{x\to0}\frac{\sin(x)}{x}$, but I was told that this is circular, because we use this limit to show $\displaystyle\frac{\mathrm d}{\mathrm dx}\sin(x) = \cos(x)$.

Do we have to use this limit to find the derivative of $\sin(x)$, or is there a legitimate counter-argument here?

Did
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Alec
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    Yes. "do I have a legitimate counter-argument here?" Dunno, depends on the argument, since you gave none it is difficult to say... – Did Sep 24 '16 at 16:05
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    @Furrer True, and offtopic: $\sin x\to0$ and, well, $x\to0$ when $x\to0$. – Did Sep 24 '16 at 16:05
  • @Did - Sorry, rephrased. – Alec Sep 24 '16 at 16:06
  • @Alec "Do we have to use this limit to find the derivative [...]" No, look on the right panel of the website, "Related" tab. E.g., this one. – Clement C. Sep 24 '16 at 16:06
  • @Furrer - Yes, and they do in this case. – Alec Sep 24 '16 at 16:06
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    Re the rephrasing: the fact that $\frac{\sin x-\sin0}{x-0}\to(\sin')(0)=\cos0=1$ seems very much like the definition of the derivative, wouldn't you say? – Did Sep 24 '16 at 16:07
  • (I am not clear on the phrasing: @Alec, do you mean "do we have to use the limit $\frac{\sin x}{x}$ at $0$ to find the derivative of $\sin$ at $0$" (well, the two are the same), or "do we have to [use L'Hopital to] find this limit" (in which case, no)? – Clement C. Sep 24 '16 at 16:08
  • @amWhy - That's ok. There are people who have evidently understood the question, and provided good comments and answers, so I don't think any elaboration is necessary. – Alec Sep 24 '16 at 16:22
  • Yes it is, assuming sine is defined geometrically and not through a power series or something like that – Anshuman Agrawal Jul 19 '22 at 15:01

2 Answers2

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the derivative of $sin(x)$ could be obtained by using the Euler notation. $sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

which gives the cosinus.

hamam_Abdallah
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It is not necessarily circular.

In the more "rigorous" books, $\sin(x)$ is defined as a power series definition, but certainly the derivation, or the way to get to such a power series depends on the aforementioned limit.

Nonetheless, purely rigorously, you do not need this limit to get the derivative, and thus you are totally allowed to use L'Hopital's Rule.

Hasan Saad
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    If I define $\sin x$ and $\cos x$ by their corresponding power series, note that those Conditions are satisfied, we could easily see that $\sin' x=\cos x$ and that $\cos 0=1$

     – Hasan Saad Sep 24 '16 at 16:18
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    The circularity doesn't have anything to do with $\sin x$ in particular. If you use L'Hopital's rule on any limit of the form $\lim_{x \to 0} \frac{f(x)}{x}$, it tells you that it's equal to $f'(0)$. But, by definition, $f'(0)=\lim_{x \to 0} \frac{f(x)}{x}$. If you already know $f'(0)$ (as you do if you're using the power series definition for $\sin x$) you don't need L'Hopital; if you don't already know $f'(0)$, L'Hopital doesn't help... – Micah Sep 24 '16 at 16:48
  • Yes, it did not help as you stated (since I already know it), but it was not a question of whether it was helpful or not. It was a matter of mathematical correctness (or otherwise). – Hasan Saad Sep 24 '16 at 16:50
  • Noticing that $f'(0)$ is equal to the limit definition we all know of seems to be irrelevant to whether we can use L'hopital or not. – Hasan Saad Sep 24 '16 at 16:51
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    In that case, when you say that it is "not necessarily circular", you are using a very different definition of "circular" than I would. – Micah Sep 24 '16 at 16:51
  • I may have a problem with mathematical terminology (as I am a foreign student and basically a self-learner of mathematics), but as far as I understand circular, it is when you use $p$ to prove $q$ when you already need $q$ to prove $p$. Isn't that what it is? – Hasan Saad Sep 24 '16 at 16:52
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    I would say that any argument whose premise is identical to its conclusion (that is, where you use $p$ to prove $p$) is circular — there's no need to identify some intermediate proposition $q$ (though you generally can). And when you apply L'Hopital's rule to a limit of this form, that's exactly what you're doing... – Micah Sep 24 '16 at 16:57
  • Perhaps, I just don't see it as circular or logically wrong. But perhaps I will learn later. Should I delete my answer? – Hasan Saad Sep 24 '16 at 16:58