The first remarkable limit is the limit of the function sin(x) / x at the point x = 0. Proof of existence and calculation of the value of this limit is necessary to find derivatives of trigonometric functions. It is stated that the limit is equal to one (this is true if the argument is given in radians; using other units, the limit value will be different). It is usually considered at the beginning of the course in mathematical analysis and its calculation is performed by comparing the area of ​​a sector of a circle and the areas of two triangles - the one inscribed in the sector and its complement to a right-angled triangle. The areas of the triangles are equal to half the sine and tangent of the angle of the sector x. With reference to the school geometry course, it is stated that the area of ​​a sector of a circle of unit radius is equal to half the value of the angle:

And this is a problem, since the formula for the area of ​​a sector includes the area of ​​a circle, the calculation of which reduces to calculating the limit of the sequence of areas of regular 2ⁿ-gons inscribed in a circle of unit radius:

And this is a special case of sin(x) / x, angles are measured in fractions of a straight angle.

Let's put aside the question of the independence of the result from the method of filling the circle with polygons (for example, you can start not with a square, but with a triangle, but there are many other ways) and try not to use knowledge of the limit. Let the area of ​a circle of unit radius be equal to Sc and the value of the straight angle be equal to P. Then the area of ​​the sector is equal to

and the existence of the limit is proved in exactly the same way, and the limit itself depends on the unknown value of the area and the arbitrarily chosen value of the straight angle:

If we choose P numerically equal to Sc, the limit will be equal to one. And by the way, the ratio of the areas of the inscribed triangle and the sector

allows one to easily prove the continuity of the sine functiom at the point x = 0. This is usually done, but the inequality is written as

I am not the first to notice this, in particular, in the article by I.I. Astakhova and V.A. Ivanov "Paradoxes of the first remarkable limit", published in the journal "Mathematical Education" (#3(63) July-September 2012) this is indicated. The article lists three options for solving the problem by defining the functions sine and cosine as:

• sums of power series (i.e. Taylor series)
• solutions of a second-order differential equation (the equation of small oscillations of a pendulum)
• functions with properties equivalent to formulas for sums of angles and sums of squares of values

The article also shows that the limit of the ratio of the length of an arc of a circle to the corresponding chord (instead of the ratio of areas) tends to one as the angle decreases. The length of a sector is calculated by integration, the integral itself contains the function arcsin(x), but this is not a problem, since the limit is calculated by L'Hôpital's rule (as a ratio of derivatives). The calculation requires knowledge of integral calculus and its application to determining the length of a curve.

It is possible to do a similar calculation of the ratio of the areas of a sector and an inscribed triangle. This is even easier, since no need to know anything about calculating the lengths of curves.

The limits of the ratios of the arc length to the chord and the area of ​​the sector to the area of ​​the inscribed triangle are equivalent to the first remarkable limit.

There was also a more important article by Yu. I. Lyubich, "Two Remarkable Limits", published in the journal "Mathematical Education" (Moscow: MCCME, 2000, Issue 4). It gave a correct proof, which I understood only after it gave me the following idea. From

• the existence of a limit of the sequence {sn} of areas of inscribed regular 2ⁿ-gons and
• strict monotonicity (decreasing) of the function sin(x) / x for small values ​​of the argument

follows the existence of the limit of the function sin(x) / x at the point x = 0 and the equality of this limit to the limit of the sequence {sn}. The monotonicity of sin(x) / x must be proven and this is the most difficult part of the matter, but for now we consider it done.

Everywhere below it is assumed that the angle is measured in fractions of a straight angle, not in radians.

Let's denote the limit of the sequence by the symbol π (this is the number Pi).

For any x > 0, the value of sin(x) / x is strictly less than π. If this were not so, there would be elements of {sn} exceeding π. This follows from monotonicity - if at the point x1 the value of the function is greater than or equal to π, one can choose a number n1 so large that

If we assume that the limit of a function does not exist or is different from π, then there exists a number m > 0 and arbitrarily small numbers x such that

Let's choose a number n2 so large that

Due to the convergence of the sequence to π this is possible. Let's choose such a number

This is possible by virtue of the assumption. But due to the monotonicity of the function

That is, the assumption is incorrect and the function has a limit equal to π.

This is all well and good, but we need to prove the monotonicity of the function sin(x) / x. For arguments of the form 1 / 2n monotonicity exists, by direct calculation we can prove that it exists for arguments of the form k / 2n. For this, we will also need to prove by direct calculation that tg(k / 2n) ≥ k tg(1 / 2n).

The assumption of non-monotonicity of the function leads to a contradiction. Let us assume that there are points 0 < x1 < x2, that in them

Let's choose such good approximations of the numbers x1 and x2 by binary fractions x1 < k1 / 2n < k2 / 2n < x2, that the values ​​of the function in them differ from the values ​​at the points x1 and x2 themselves by less than m / 2. But then we get

and this contradicts the monotonicity of the function for arguments of the form k / 2n.

The equality sin(x1) / x1 = sin(x2) / x2 also impossible. If we assume it, then by choosing two points x1 < k1 / 2n < k2 / 2n < x2 we get at the case of inequality, since at these selected points there is no equality of function values.

If sin(k1 / 2n) / (k1 / 2n) ≤ sin(x1) / x1, then sin(k2 / 2n) / (k2 / 2n) < sin(x2) / x2, and instead of the point x1 we need to take k2 / 2n.

Otherwise, if sin(k1 / 2n) / (k1 / 2n) > sin(x1) / x1, then instead of the point x2 we need to take k1 / 2n.

In Lyubitsch's article, the concavity (upward convexity) of the function sin(x) and the zero value at zero argument are used to prove monotonicity. To prove the concavity itself, the formula for the sum of the sines of two angles and the continuity of the sine are used. This is shorter, but you need to know about convexity-concavity.

The proof is based on the continuity of the sine function, the proof of which is based on knowledge of the formulas for the sum and difference of angles and is reduced to proving the continuity of the sine function at the point x = 0. Continuity follows from the monotonicity of the sine function in the vicinity of the point x = 0 and the existence of converging to zero sequence of sine values ​​at the points xn = 1 / 2n. That is, here too, no comparisons of the values ​​of the sine function and its argument are needed.