A Limit That Converges to Pi- Mathematical Explanation
What Limit Actually Converges to Pi?
Most people know that π shows up in circles. Fewer know that π appears in limits—mathematical expressions that approach a value as you push them toward infinity. The most elegant one is:
n · sin(π/n) as n → ∞ equals π
That limit isn't just a curiosity. It's a bridge between trigonometry and infinite processes. Here's why it works.
The Setup: Why This Limit Exists
Start with the function f(n) = n · sin(π/n). As n grows, two things happen simultaneously:
- sin(π/n) shrinks toward 0
- n grows toward infinity
- The product stabilizes around 3.14159...
This is an indeterminate form of type ∞ · 0. The limit exists because the rate at which sin(π/n) shrinks beats the rate at which n grows.
The Proof: Using the Squeeze Theorem
You need one key fact: for small x (in radians), sin(x) < x < tan(x). This geometric truth comes straight from the unit circle.
Step 1: Set Up the Inequality
For any positive x:
sin(x) < x < tan(x) = sin(x)/cos(x)
Rearrange the right side:
cos(x) < sin(x)/x < 1
Step 2: Substitute x = π/n
When n → ∞, π/n becomes tiny. Substitute:
cos(π/n) < sin(π/n) / (π/n) < 1
Multiply everything by π:
π · cos(π/n) < n · sin(π/n) < π
Step 3: Apply the Squeeze
As n → ∞, cos(π/n) → cos(0) = 1. The lower bound squeezes up to π. The upper bound stays π.
The result: n · sin(π/n) → π
That's it. The squeeze theorem delivers the answer with minimal algebra.
Intuition: Why This Makes Sense
Picture the unit circle. Take a tiny angle θ = π/n. The arc length is θ. The chord length (straight line) is 2·sin(θ/2).
For small angles, the chord and arc are nearly identical. As n gets huge, n·sin(π/n) approximates the arc length of a half-circle—which is exactly πr, and with r=1, that's π.
The limit is essentially measuring half the circumference of a unit circle through increasingly fine approximations.
Other Limits That Give Pi
The n·sin(π/n) limit isn't alone. Several other expressions converge to π:
| Limit Expression | Convergence Speed | Notes |
|---|---|---|
| n·sin(π/n) | Slow | Clean proof via squeeze theorem |
| (2^(2n)·(n!)²)/((2n)!·√n) | Moderate | Related to Stirling's approximation |
| 4·∑(-1)^k/(2k+1) | Very slow | Leibniz series (alternating) |
| ∫₀¹ 4/(1+x²) dx | Instant | Direct integral definition |
The integral definition is technically a limit (as the partition gets finer), but the n·sin(π/n) form is cleaner for understanding convergence behavior.
How to Use This in Practice
Step 1: Pick a value for n. Start small—n = 6 gives 6·sin(π/6) = 6·0.5 = 3.0
Step 2: Increase n. n = 10 gives 10·sin(π/10) ≈ 3.09
Step 3: Watch convergence. n = 100 gives ≈ 3.1395. n = 1000 gives ≈ 3.14157
Step 4: Verify. The true value is 3.14159265...
By n = 1000, you're accurate to 4 decimal places. The error drops roughly as 1/n².
Why This Matters
This limit isn't just textbook material. It appears in:
- Riemann sum derivations of circular area formulas
- Probability through normal distribution approximations
- Signal processing where sinc functions dominate
The sinc function is defined as sin(x)/x. The limit n·sin(π/n) = π is really a scaled version of the fundamental limit sin(x)/x → 1 as x → 0.
Master this limit and you understand the backbone of half the calculus you'll encounter in engineering and physics.
The Bottom Line
n·sin(π/n) converges to π because sin(π/n) behaves like π/n for large n—the ratio sin(x)/x approaches 1. Multiply back: n·sin(π/n) approaches π.
No motivational quotes. No "journey of discovery." Just math that works, because the geometry demands it.