10
$\begingroup$

If $\frac{d(\sin{x})}{dx}= \cos{x}$ is proven using the limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (as it is in most textbooks), would it be circular to then use $\frac{d(\sin{x})}{dx}= {\cos{x}}$ and L'Hospital to prove the limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (because limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}= \frac{\cos{0}}{1})$ later in the same textbook?

  • 11
    Yes indeed it would be circular.2012-03-06
  • 0
    What if, instead of using d(sinx)/dx=cosx you argued d(sin)/dx at zero = lim,h->0 [sin(0+h)-sin(0)]/h and, by Taylor series, sin(0+h)~h, so lim,h->0 =h/h =1 ? Is this a valid alternative?2012-03-06
  • 0
    Can you explain the aversion to the standard proof? It doesn't even REALLY use $\epsilon$-$\delta$s, it squeezes the function between $\cos$ and $1$.2012-03-06
  • 3
    @johnthemathteacher: And how are you going to establish that the Taylor series of $\sin(0+h)$ is $\sim h$ without knowing what the derivative of $\sin(x)$ is, or without computing $\lim_{h\to 0}\frac{\sin(0+h)}{h}$?2012-03-06
  • 6
    What you *can* do is: once you have *established* that $\frac{d}{dx}\sin x = \cos x$, if you are having trouble remembering what $\lim\limits_{h\to 0}\frac{\sin h}{h}$ *is*, then you can use L'Hopital's rule to *remind* you of the value, though not to *prove* its value ex nihilo.2012-03-06
  • 0
    @john the mathteacher: In a few presentations of the calculus, series more or less come first, then the sine function is **defined** in terms of its power series. Or else differential equations come first, and the trig functions are defined as solutions of the usual $y''=-y$, and their basic properties are derived therefrom. But in conventional development, for the series for sine we need its derivative, so the argument is circular, not much more subtly so than the L'Hospital Rule method.2012-03-06
  • 0
    IF, sine is **defined** by the power series, then is my argument above correct? "lim,h->0 [sin(0+h)-sin(0)]/h and, by **the definition of sine as the power series h-h^3/6+...**, sin(0+h)~h, so lim,h->0 =h/h =1"2012-03-06
  • 1
    @john the mathteacher: Yes, that would be fine. The derivation of the basic properties is technically rather painful, from the series you get the DE, then $\sin^2 x+\cos^2 x=1$. Then one proves $\cos$ is periodic, $\pi$ gets defined in terms of the period. Not for a first course!2012-03-06
  • 0
    Thanks for the responses. I asked this question because someone claimed that proving lim,x->0 (sinx)/x=1, in an informal discussion using, L'Hospital is circular. Based on these responses I would say that it is only circular if that particular proof of d(sinx)/dx=cosx has been used previously. Would you say that that one is justified in proving it from L'Hospital without reference to how d(sinx)/dx=cosx has been proven (because there are non-circular proofs of d(sinx)/dx=cosx)? Or is the argument circular because the assumed proof of d(sin)/dx =cosx is the elementary, but circular, one?2012-03-06
  • 3
    @johnthemathteacher: Since the vast majority of the modern developments do not start from the power series, for the vast majority of developments using LHopital's rule *would* be circular; in order to *justify* that the LHopital's rule argument is not circular you would likely have to specify that you are defining $\sin x$ as a power series, and deriving its properties from that starting point. Otherwise, for *most* people, the argument *will* be circular. I would not be satisfied with an **unspoken** "and it is not necessarily circular because there is other roundabout way of proving this".2012-03-06
  • 0
    I didn't realize that lim,x->0 sinx/x was that fundamental.I thought there would be other elementary proofs of d(sinx)/dx=cosx. I suppose the geometrical proofs aren't rigorous enough. Thank you.2012-03-06
  • 0
    As I recall, there are complete geometric proofs invoking the squeeze/sandwich theorem that $\sin x/x \to 1$.2012-03-06
  • 6
    Why is it that no one has pointed out the pun involved in using "circular" to describe a problem about $\sin x$?2012-03-07
  • 0
    @Gerry It did baffle me when someone wrote "Circular definition [for the sine]..." and explained how we can defined the sine in terms of the unit circle. I thought: "Why would you say it is circular?" Then I went "Ohh... **circle**"2012-03-07

1 Answers 1