The limit of limiting arguments

A link to the full video is at the bottom of the screen.

That video gives multiple examples of lying with visual proofs

Editing from the original video into this short by Dawid Kołodziej

[matthewb]

This is a perfect approach for approximating the AREA of the circle. Indeed, calculate the area of that region, and you’ll find it approaches pi. But the perimeter length does not approach the circumference length just because the boundaries of the shapes get arbitrarily close to each other. This is because one is infinitely jagged.

[QobelD]

Here's the correct answer: The perimeter of the shape being altered, to more closely match the perimeter of the circle, never changes in size, so the limit being approached is not the circle's perimeter. What is changing is the area between the corners of the shape enclosing the circle and the perimeter of the circle. This means the limit being approached is either the area of the circle, or the least area between the perimeter of a shape that can enclose a circle, and the circle it encloses. The limit is either pi or 0, respectively.

[leftylizard]

Basically, in order for an apporximation to work, it has to actually approach the value you want rather than staying the same forever, no matter how many times you repeat the process.

This actually does work for approximating the area of the circle because with each new iteration, some area is lost, and so it's value changes, shrinking closer and closer from 4 to pi as you go on.

Meanwhile, the perimeter gets stuck on 8 forever.

[andreochimal]

For anyone looking for the real formal answer, the problem is that arc length is only lower semicontinuous, and this approximation of length is always greater than the length itself (upper, not lower), so you cannot exchange the limit with the function, i.e. the length of the limit curve is not the limit of the lengths. However, if you approximate with a curve that tends towards a circle but with a smaller perimeter you would be able to interchange the limit and the function

[jonathanaarhus]

This is why the Squeeze Theorem is so important.

[curtfehr]

Limits can’t be used to calculate the curves themselves, only the area under the curves. If you really wanted to approach the distance of the curve you would have to calculate and add up all the hypotenuse’s between each point.

[elliotgregory]

What an amazing demonstration of visual proofs, and how they can be used to say whatever you want. Keep up the good work 3b1b team, and merry christmas ❤

[albertotonon]

Is It because the square curve is not differentiable? You can make the shape go infinity small, but you end up with a fractal

[HRDBMW]

This one blew my mind until I suddenly got it.

Thank you for the puzzle!

[robmac]

The same argument shows that the square root of 2 is 2. Take a 45, 45, 90 triangle, say defined by the points (1, 0), (0, 0), and (0.1). Walking from (1, 0) to (0, 1) directly is length square root of 2. Walking to (0, 0) then (0, 1) is length 2. Now, take the jagged, sawtooth path with more and more horizontal jigs and vertical jags...still has length 2 but gets closer to

[RandomUser]

This same algorithm with proper adjustment can actually be applied to any shape to argue it’s perimeter is that of an appropriately sized square from which would fold the corners.

[johnchessant]

The problem is that even in the limit, you can think of it as going around the two legs of an infinitesimal right triangle rather than the hypotenuse. More formally, we say that even though this shape approaches a circle, the velocity vector as you traverse the shape does not approach the velocity vector as you traverse the circle.

[thebookkeeper.k]

So one time in 10th grade I did a project on this thing called taxicab geometry. I would definitely recommend looking into it if you’re bored. It’s basically like Minecraft water mechanics. In taxicab pi does actually equal 4 without breaking any rules of calculus

[petachad]

Because in a limit, with each iteration, the value changes and converges to a finite value(if the limit is converging), but in this case, for each iteration the value remains the same(i.e 8) and the limit of a constant is the constant itself.

[TJStellmach]

We can show _that_ it's not okay (without really saying why not) by observing that we can do essentially the same thing with an inscribed octagon, which has a known perimeter not equal to 4.

[Solarch-wcvs]

You can modify the square infinitely to look more like the circumference but it will never be the same. Eventually each section of the circumference, that is so small that it looks completely flat, becomes the hypothenuse on a right triangle with the 2 shorter sides being the part of the square that is supossedly the same. For pie to be 4, you'd be saying that the hypothenuse is equal to the sum of the shorter sides in a right triangle and that's just not even close.
Even if the square has been transformed to be identical to the circumference at plain sight, it's still a zigzaging line that is in reality much longer. Same way your intestines are a certain length inside the body but their effective area is much bigger than expected because of all the microscopic folds they have on their surface

[mehrdadmohajer]

Thx. 1) you can Approach from outside Boundary toward the Circle & that's no Problem.
2) It is The Tangent- Line which corresponds to Pi at Point of Tangency.
Cheers🍻

[glennallan]

I have had this thought also. Good to see that someone put it graphically. <3

[michaelkelly]

So many comments about nothing. It is actually very easy to see. Each approximate line must be tangent in all points around the circle. It is tangent only in 4 points which is clearly wrong.

[nickschneider]

Because for triangles A^2 + B^2 = C^2, and the limit as A->0, B->0 does not approach A+B=C continuously