It is a very well-known result in mathematical literature and proven in almost all analysis undergraduate books and lectures that

The proof essentially goes as follows.

Let . Let be a line and let be a point on it. Draw a sector with as the center and an angle radians, where is on . Drop the perpendicular from to and call it . Extend and let the tangent to the arc at intersect this extension at . A picture of this is as follows.

The proof then proceeds by noting that

which gives

Simplifying, we get

Then, appealing to the fact that is right-continuous at 0, the squeeze theorem is applied to conclude that

Similarly, flipping the diagram and using the fact that is left-continuous at 0, we get

Combining these two results finishes the proof.

There are a couple of small and a large problem with the above proof that are generally not addressed in most books.

1. The fact that the area of is less than or equal to the area of the sector , while clear from the picture, requires a formal proof. Given that 'area' satisfies , it suffices to show that the triangle is a subset of the sector (why area satisfies that property comes from the area being the Lebesgue measure on ). That is very easily done.

2. Similarly, although a bit less of an issue, is that no argument is generally given for showing that the area of the sector is less than or equal to the area of and is just asked to refer to the picture. We shall give a proof for this as well.

3. The continuity of is also a small issue that is generally taken for granted. While intuitively clear, one does need a proof for it. It easily follows from the setup, and we will see how one goes about doing it.

4. The bigger problem is the claim that the area of the sector is . This is generally shown by saying that the area of the circle is , so an angle of radians cuts off a sector of area . Since the radius is 1 here, the result follows. But how do we know that the area of the circle is ? Well, the go-to method is to prove it via integrals.

   • A standard proof goes as follows.

     The circle of radius is given by the equation . So, the area is given by

     But here we have used that the derivative of is whose proof also uses the limit in question. Thus this proof for the area of a circle cannot be used.

   • There is also another proof that tries to avoid differentiating any trigonometric function. This is the so-called shell method, where we divide the circle into very small rings (infinitesimally small) of radius and thickness . These are of area and integrate over them to get

     But this also hides the derivatives when looking through the lens of rigor. Where are the derivatives? Well, the shell method is rigorously proven by the change-of-variables formula. In this case, we change variables from to (cartesian to polar). The Jacobian of this transformation is where the derivatives are hidden.

   • How about a non-integral proof? Like say Archimedes' proof, where he shows that the area of the circle is neither greater than nor less than the area of a triangle with a base equal to the circumference and height equal to the radius?

     If one reads very carefully, the proof assumes that if a curve converges pointwise to another curve (in this case, inscribed polygons converging to a circle), then so does the areas. This is not rigorously justified and can fall prey to the staircase paradox. The rigorous justification again uses the limit in question so... we get the point.

As we argued before, it is enough to show that is a subset of the sector . It will follow from some very simple results.

The main idea would be to show that triangles and sectors are convex, and that triangles are the smallest convex set defined by any 3 points. And then we will note that we have our triangle and sector defined by the same set of points, and hence the sector must be at least as big as the triangle.

By construction, is perpendicular to at , and lies on the line joining and . Thus, the line joining is a tangent to the circle at (this follows from basic geometry). If we can show that a tangent of a circle does not intersect the interior of a circle, then must be on the extension of towards , and we will have that the sector is a subset of , and we will be done.

To that end, suppose we have a circle with the center at and a line tangent to the circle at . By basic geometry, we know that is perpendicular to the tangent line. Suppose the tangent line intersects the interior of the circle at some point . Then since is on the circle and is in the interior. We look at . Clearly
and thus a contradiction. Thus, a tangent always lies outside the circle except at the point of tangency.

This completes the proof for this section.

This is easier than it seems. We first note that by the result in section 2, we have that for . Using the squeeze theorem, we have that . Now we note that . Now note that this implies that cosine is even. Thus, the limit of sine from one side is enough to guarantee cosine's limit from both sides and we have that . Thus for continuity, we check that
and hence cosine is continuous at 0.

This is the trickiest of the mentioned problems. We already saw several circular proofs of this. Can we resolve this?

Let us look at our goal. We want to find the area of the circle without ever using trigonometric functions or unjustified limits. We'll actually use calculus to do this. Let us set everything up.

For our circle, the equation is . Well, the function is branched so that's an issue, but we can fix that by noting that if we can find the arclength of the upper semicircle (the positive branch), then by symmetry, the whole circle has arclength twice of that. So we only consider . Then . Plugging this in, and going from to , we get (letting be this half circumference)

Note that the integral is independent of any circle we draw, and thus this is a genuine constant. Calling this (yes, that ), we have the result that . The other half of the circle also gives another , and thus we have the grand result that the circumference of a circle with radius is .

How about the area now? Well, let the area be . Basic calculus tells us that the area of the top semicircle is

We can massage this into an expression that is in terms of and that will allow us to conclude our result. So lo and behold...

Proof.
As we said, the proof will be a delicate massaging of the above integral

Thus we have that □

Thus the area of the whole circle is . This completes the proof for the area of the circle. Note that no derivatives of trigonometric functions went into calculating this (there is obviously trigonometry hidden when we write the equation of a circle, but there's no derivatives involved.)

Yeah, I am too lazy to write a conclusion. This took way more time to finish than I was expecting because I am apparently a top-tier procrastinator. Anyway, we showed that the limit of as is 1 with almost all the rigor necessary. If something is still missing, hit me up I guess?