4D geometric and limit of (x - sin x)/x³

[hy.dodec]

sinxx.png (8.28 KiB) Viewed 164 times

In this picture, there're two right triangles and a circular sector whose radius is 1.
If we rotate these shapes by x-axis, then two right triangles become two cones and a sector becomes a spherical sector.
We can evaluate the limit of (1 - cos θ)/θ², θ → 0+ by comparing volume of each shape.

Volume of each shape can be evaulated by integrating areas of circular cross sections, but what would happen if we integrate volumes of spheres instead here?
Then two cones become two spherones and a spherical sector becomes something 4D shape.

Let's evaulate hypervolume of these 4D shapes:

극한.png (16.29 KiB) Viewed 163 times

Finally we can evaluate the limit of (x - sin x)/x³, x → 0+ by comparing these hypervolumes. Multiply 4/π to all sides then subsitute 2θ = x, divide all sides by x³, apply squeeze theorem.

For x → 0-, just substitute s = -x then you'll got same result of x → 0+.

[mr_e_man]

something 4D shape

How about "hyperspherical sector"?


I glanced at your picture and volume formulas, and then thought about it later, but I misremembered the picture. You drew a tangent from A to the line OB. In my mind I drew a tangent from B to the line OA. So I got a different volume formula for the larger shape:

π/3 sin²θ tan θ

Anyway, it gives the same limit.

[mr_e_man]

Here's a more algebraic way to find the limit, assuming it exists:

L = lim (x - sin x)/x³
= lim (2θ - sin 2θ)/(2θ)³
= lim (2θ - 2 sin θ cos θ)/(8θ³)
= 1/4 lim (θ - sin θ cos θ)/θ³
= 1/4 lim (θ - θ cos θ + θ cos θ - sin θ cos θ)/θ³
= 1/4 lim (θ(1 - cos θ) + (θ - sin θ)cos θ)/θ³
= 1/4 lim (1 - cos θ)/θ² + 1/4 (lim (θ - sin θ)/θ³) (lim cos θ)
= 1/4 (1/2) + 1/4 (L) (1)
= 1/8 + L/4

8L = 8(1/8 + L/4) = 1 + 2L
6L = 1
L = 1/6

[hy.dodec]

something 4D shape

How about "hyperspherical sector"?

I have guessed that too, but I'm not sure that's correct

Here's a more algebraic way to find the limit, assuming it exists:

(...)

Here's another algebrical solution.
TMI ) I think this solution is best way for high school students whose curriculum doesn't include L'Hôpital's rule and taylor series.

We can get below inequality from volumes of two cones and a sector that mentioned above:


Since sec θ > 1 in the range (0, pi/2), the following inequality is also true.


all three functions passes Origin, this inequality still holds after we integrate all sides in [0, θ] (0 < θ < pi/2)
So we get another ineqauilty:


Divide all sides by θ³, then apply squeeze theorem.