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$$\lim\limits_{x \to 0} \frac{\sin x}{x}= \cos 0 = 1$$

This result follows from either L'Hospital's rule or by definition of derivative of $\sin x$ at $x=0$. I see widespread accusation that finding the limit this way is circular reasoning. However, I can hardly agree with it. Otherwise, you may as well accuse that summing an Infinite Series by Definite Integral is circular.

In my humble opinion, relating a limit back to the derivative or integral of a function should be a perfectly legitimate (and sometimes very clever) trick. Why is it accused as circular reasoning by so many people?

Furthermore, it seems that a majority of people making this accusation fail to provide a better way to find the limit concerned. Many of their attempts assume the area of a sector to be known. I don't see how assuming the area of sector to be $x/2$ is any better than assuming the derivative of $\sin x$ to be $\cos x$.

Feel free to share your point of view on this.

Below are some discussions on the limit concerned. https://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Richman160-162.pdf http://forums.xkcd.com/viewtopic.php?t=112236

José Carlos Santos
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Stupid Girl
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    How do you prove that $\frac{d}{dx} \sin(x)=\cos(x)$? – Harto Saarinen Sep 24 '17 at 06:37
  • @HartoSaarinen power series? – Epiousios Sep 24 '17 at 06:38
  • It depends on how you define sin x. First principle is not always required. – Stupid Girl Sep 24 '17 at 06:39
  • Try to prove it from the definition of derivative. And see what happens. That's the origin for the claim that there is circular argument. – Harto Saarinen Sep 24 '17 at 06:39
  • I think showing the area of a unit sector to be x/2 through other means is not any easier than showing the derivative of sin x to be cos x through other means. – Stupid Girl Sep 24 '17 at 06:42
  • It really all depends how you define the trigonometric functions. And I didnt say it would be easier, Im just explaining why the argument is many times looked to be circular. – Harto Saarinen Sep 24 '17 at 06:44
  • The answer depends on the context the "accusation" is made in, whether a deep discussion on the axiomatic foundations of mathematics, or a student arguing on their exam marks. In the latter case, there is a limit to how far you can establish the premises. – Miguel Sep 24 '17 at 06:46
  • Don't worry about exam marks. In our exams, we are expected to recite this limit and give the answer straight away (with no intermediate step). – Stupid Girl Sep 24 '17 at 06:51
  • Could you also share your point of view on summing an Infinite Series by Definite Integral? Is it circular reasoning or a legitimate trick? https://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/ – Stupid Girl Sep 24 '17 at 06:53
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    The fundamental problem here is not the use of derivative of $\sin x$ but rather the use of L'Hospital's Rule. If by any means you know that derivative of $\sin x$ is $\cos x$ then a trivial/immediate corollary is that the above limit is $1$. You dont need L'Hospital's Rule to find this limit. There is no circularity involved here anyway. See this answer: https://math.stackexchange.com/a/2277563/72031 – Paramanand Singh Sep 24 '17 at 09:19
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    Summing an infinite series via integral (as given in your link) is not a trick, but rather the definition of integral (actually its an immediate corollary of the definition). The power of the method is not the integral representation but rather the use of Fundamental Theorem of Calculus to evaluate the integral afterwards. If the Fundamental Theorem of Calculus is not available the integral representation does not help at all in evaluating such infinite sums. – Paramanand Singh Sep 24 '17 at 09:31
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    Also in geometric definition one does not prove that sector of unit circle has area $x/2$, rather one says that if the area of sector is $t$ then by definition the coordinates of the point are $(\cos 2t,\sin 2t)$. – Paramanand Singh Sep 24 '17 at 09:34
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    A derivative is always evaluated via first principle (first principles is nothing but definition of derivative). But the procedure is not always direct. Rather first principles are used to prove rules of derivatives which can then be used to evaluate the derivative very quickly /efficiently. Remember that when we have to sum first 100 natural numbers we don't perform 99 additions, but rather use the formula for sum of arithmetic progression (which is derived using properties of addition and multiplication). – Paramanand Singh Sep 24 '17 at 09:38
  • Have a look at https://math.stackexchange.com/a/2320028/44121, too. – Jack D'Aurizio Sep 24 '17 at 12:21
  • @ParamanandSingh The proof of Fundamental Theorem of Calculus would rely on the definition of the definite integral as an infinite series. Similar accusation could have been made. – Stupid Girl Sep 24 '17 at 13:22
  • @StupidGirl : the definite integral (Riemann integral to be precise) is defined as a very special kind of limit of a sum. There is no other substantially different way to define integral. Perhaps you have learnt the definition of integral as difference between values of anti-derivarive. That's total crap and it is time to change textbooks. And the proof of Fundamental Theorem of Calculus is not so difficult. There is no circularity involved here. – Paramanand Singh Sep 24 '17 at 14:20
  • @ParamanandSingh It is not quite about definition. I am just pointing out how that calculating a Riemann sum by considering its anti-derivative is no different from calculating the limit of sin x/x as x->0 by considering the derivative of sin x at x=0. – Stupid Girl Sep 24 '17 at 14:49
  • If some people accuse that the calculation of the limit (sin x)/x as x->0 should not make use of the derivative of sin x, they could as well accuse that the calculation of a Riemann sum should not make use of the FTC. – Stupid Girl Sep 24 '17 at 14:56
  • The most common technique for evaluation of an integral is using anti-derivative (via Fundamental Theorem of Calculus) and there is no circularity involved here. The case for evaluation of limit in question using derivative is something different. I don't understand how you treat both of them in the same manner. I have not heard anyone raising a finger at evaluation of integrals via FTC. Without FTC one can evaluate only very few integrals. – Paramanand Singh Sep 24 '17 at 15:46
  • https://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/ The situation I am referring is a bit different. Here, it is requested to find the limit of an infinite series (not an integral). Incidentally, it equals to the Riemann sum of a particular definite integral. Then, FTC is used to calculate that definite integral. Obviously, the value of that definite integral would be defined by the original infinite series to be found. – Stupid Girl Sep 24 '17 at 16:05
  • Just to be precise the kind of limit you are talking in that link is not an infinite series but rather a limit of the form $\lim_{n\to\infty} \sum_{k=1}^{n}f(n,k)$. An infinite series is of the form $\lim_{n\to\infty} \sum_{k=1}^{n}f(k)$. But lets use your terminology. The infinite series is not incidentally related to an integral, but rather this relation between the integral and the series is an immediate consequence of the definition of integral. And integrals are generally evaluated using FTC rather than using the definition of integral. Don't understand why one would accuse here. – Paramanand Singh Sep 24 '17 at 18:47
  • Related : https://math.stackexchange.com/questions/2118581/lhopitals-rule-and-frac-sin-xx – Arnaud D. Dec 03 '19 at 14:55

2 Answers2

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I once saw the definition of $\sin(x)$ for $x\in [-\frac{\pi}{2}, \frac{\pi}{2}]$ as the inverse function of $$\arcsin(t) = \int_0^t\frac{1}{\sqrt{1-u^2}} du$$ It follows that $$\sin^\prime(x) = \sqrt{1-\sin^2(x)}$$ hence $\sin^\prime(0) = 1$ and it follows that $\lim_\limits{x \to 0} \frac{\sin(x)}{x} = 1$.

Epiousios
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Gribouillis
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  • How would you derive the integral represention of the arcsine function without knowing the derivative of sine imir arcsine? – MrYouMath Sep 24 '17 at 07:03
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    You don't derive the integral representation of arcsine. You define arcsine by the above integral, in the same way that you define $\log(x) = \int_1^x\frac{d t}{t}$. – Gribouillis Sep 24 '17 at 07:05
  • Such procedures make no sense. You can’t define the inverse of a function to be something. Then you could also define sine of x as the integral of cosine x. – MrYouMath Sep 24 '17 at 07:08
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    @MrYouMath The difference is that $1/\sqrt{1-x^2}$ is a well known function defined by elementary means. This construction allows one to define trigonometric functions from polynomials and square roots. In this presentation, sine is defined as the inverse of arcsine, which definition does not need trigonometry. – Gribouillis Sep 24 '17 at 07:10
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    I agree with Gribouillis. This is a perfectly legit way of doing it. –  Sep 24 '17 at 08:00
  • The difficulty now seems to be to relate this definition back to the unit circle. – A.Γ. Sep 24 '17 at 08:43
  • @A.Γ. It is not that difficult, because the length of the curve $f(x) = \sqrt{1-x^2}$ is $\int_{-1}^1 \sqrt{1 + (f^\prime(x))^2} d x = \int_{-1}^1 (1 - x^2)^{-1/2} dx$. One only need to develop the theory from here. There could be other starting points, such as the primitive of $\frac{1}{1+x^2}$. – Gribouillis Sep 24 '17 at 08:50
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    @MrYouMath There is nothing wrong in defining $\displaystyle a(x)=\int_0^x\frac{\mathrm dt}{\sqrt{1-t^2}}$, proving that it is injective. and then to define $\sin|_{\left[-\frac\pi2,\frac\pi2\right]}$ as its inverse. – José Carlos Santos Sep 24 '17 at 08:55
  • This is the geometric definition of circular functions, but very rarely do textbooks mentions this while discussing the geometric definition of circular functions. +1 – Paramanand Singh Sep 24 '17 at 09:21
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Normally this result is needed to derive the derivative of trigonometric functions (with the geometric interpretation) in the first place. So using the derivative for the limit makes this seem circular from this point.

On the other hand if you use the taylor series as the definition of trig functions then the limit is trivial by using the series of the sine function. And in order to derive the taylor series we also first need the derivatives of sine and cosine. Hence, one could argue that this is again circular.

The geometric way of deriving the limit has also flaws but it uses very simple concept that is why I would say that it is better than using the derivative/taylor series derivation.

The only thing you have to accept for the geometric derivation is this inequality

$$\sin x\leq x \leq \tan x $$

for small positive $x$ this is very easy to see from the geometric picture of a unit circle.

MrYouMath
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    If you define trigonometric functions by their power series, you get their derivatives as a consequence, there is no circle. But you'd need all the theory, including uniform convergence of function series, before you can do that, that's a drawback. And you'd have to prove that the functions so defined have their uses in geometry, and that's not a trivial task. –  Sep 24 '17 at 06:46
  • The inequality relies on the area of sector, which is not trivial. – Stupid Girl Sep 24 '17 at 06:56
  • @ProfessorVector That is what I meant. In order to make it reasonable to use the taylor series as a definition we would first need the derivatives of cosine and sine in the first place. And as you allready said all the theory needed to work with infinite series is also necessary. – MrYouMath Sep 24 '17 at 07:01
  • @stupidGirl It can also be seen from the arclengths. Sine is inside the circle, x is the arclength of the circle and the tangent of x is tangent to the circle. The arclength x is then bounded by sine of x from below and the tangent of x from above. – MrYouMath Sep 24 '17 at 07:06
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    It is not trivial whether the arclength has to be shorter than the tangent. – Stupid Girl Sep 24 '17 at 07:15
  • Could you also share your point of view on summing an Infinite Series by Definite Integral? Is it circular reasoning or a legitimate trick? https://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/ – Stupid Girl Sep 24 '17 at 07:16
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    @StupidGirl: I would not think that the Riemann summation trick is circular reasoning. It is a reasonable definition for evaluating integrals. Without knowing the derivative and anti-derivative concept this is the only way you can evaluate an integral. – MrYouMath Sep 24 '17 at 09:10
  • @StupidGirl: the inequalities are derived from the basic fact that area of a region can never be exceeded by area of a sub-region (and the definition of circular functions). – Paramanand Singh Sep 24 '17 at 09:43
  • The treatment via arc lengths is difficult to handle using geometric arguments. The inequality $x\leq \tan x$ requires some deep arguments about convexity of the circle and it is preferable to use integrals rather than geometric arguments. – Paramanand Singh Sep 24 '17 at 09:46
  • @MrYouMath The aproach in that blog is actually evaluating an infinite series by noticing it is the Riemann sum and then using the anti-derivative to evaluate. – Stupid Girl Sep 24 '17 at 13:13
  • @ParamanandSingh I agree that the treatment via arc length is difficult. Unfortunately, radian is normally defined in terms of arc length. Also, many of those people making the concerned accusation actually fail to address the subtlety involved. Their way of finding the concerned limit does not look any better than the way they accused. – Stupid Girl Sep 24 '17 at 13:38
  • @StupidGirl: if one uses arc-length then the derivative of $\sin x$ is immediate as is the inequality in this answer. The difficulty arises only when we want to argue using geometry. The proper/simpler approach is to use integrals (arc lengths are evaluated using integrals) and then the problem is trivial. – Paramanand Singh Sep 24 '17 at 14:16