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How would you find $\lim\limits_{x\to 0} \frac{\sin x}{x}$? I have tried the every method I can think of, but I am not getting correct results.

TheLabyrinthMaker
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  • If you know the derivative of $\sin$, you can use the definition of the derivative at $0$, or L'Hôpital. If you don't know the derivative of $\sin$, you need to use a geometric argument. – Akiva Weinberger Oct 15 '15 at 01:21
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    @AkivaWeinberger How could you know the derivative of $\sin$ without knowing this limit? –  Oct 15 '15 at 01:22
  • Wow, thanks for the fast response! That definitely solves my problem, thank you. – TheLabyrinthMaker Oct 15 '15 at 01:25
  • @avid19 I dunno, maybe you forgot it or something. – Akiva Weinberger Oct 15 '15 at 01:26
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    @AkivaWeinberger The point is that using derivatives here is circular! –  Oct 15 '15 at 01:27
  • @avid19 That assumes there's only one way of proving that $\sin'=\cos$. And it's pretty irrelevant whether or not it's circular; I'm just saying that, if you know $\sin'(0)=1$, then you can derive the above limit without too much effort. – Akiva Weinberger Oct 15 '15 at 01:30
  • @AkivaWeinberger You can not use the derivative of sin in any manner to find the answer to the limit. The derivative of sin can only be found if you already know the answer to the limit to begin with. To state that it is "pretty irrelevant whether or not it's circular" is rather astounding... – imranfat Oct 15 '15 at 02:02
  • @imranfat What I mean is:$$\sin'(0)\implies\lim_{x\to0}\dfrac{\sin x}x=1$$is easier to prove than:$$\lim_{x\to0}\dfrac{\sin x}x=1$$ – Akiva Weinberger Oct 15 '15 at 02:39
  • @AkivaWeinberger True! But I would like to know, how do you know that $sin'(0)=1$ without knowledge of the derivative of sine? Because then this would be a nice alternative to prove the limit in question... – imranfat Oct 15 '15 at 02:47
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    @avid Define the sine function as the inverse of $\int_0^x \frac{1}{\sqrt{1-x^2}},dx$. Then, one can easily show that $\frac{d\sin x}{dx}=\sqrt{1-\sin^2 x}$. And evaluation at $x=0$ is trivial. So, yes, it can be done with standard theorems and circumvents the circular logic. – Mark Viola Oct 15 '15 at 03:44
  • @imranfat See my previous comment for a way forward that avoids circular logic. – Mark Viola Oct 15 '15 at 03:47
  • @Dr.MV I see what you are aiming at. Certainly interesting. I do find this "definition" a little troublesome though as the sine already is defined in a particular way in the realm of trigonometry. The inverse follows then as a consequence and so there is nothing to define. With this kind of "definition", does it follow that its inverse is equal to the y-coordinate of a point on the unit circle, which is what the sine really is? – imranfat Oct 15 '15 at 13:51
  • @imranfat Yes, with that definition one can prove equality with the circular function interpretation. But I disagree that "is what the sine really is." A more mature definition is, for example, $\sin z =\sum_{n=0}^\infty \frac{(-1)^{n}z^{2n+1}}{(2n+1)!}$ for all complex $z$. – Mark Viola Oct 15 '15 at 13:57
  • @Dr.MV I see what you are saying, except, the sine already existed at the time of Hipparchus, by many seen as the founder of trig. Taylor series came around at the time of Newton :) – imranfat Oct 15 '15 at 15:57
  • @imranfat That is a fair point. Nevertheless, the approach I suggested is still a rigorous way forward. – Mark Viola Oct 15 '15 at 16:20

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