0^0

[Albert Chan]

(11-24-2023 08:14 PM)klesl Wrote:  I agree with Werner, 0^0 is undefined with limit equals to 1 as x approaches to 0

0^0 is undefined because there is *no* 1 value for limit, we can make it to anything.
What Werner is saying is many f^g cases (f→0, g→0), the limit is 1
For convenience, we may define 0^0=1, but this is not the truth.

Limit depends on relative rate of base and exponent shrink to 0
For example, we have identity: ε ^ (1/ln(ε)) = e

Make exponent shrink to zero a tiny bit slower, we get 0^0=0
Cas> limit(ε ^ (1/(-ln(ε))^0.99999), ε, 0, 1)      → 0

Or course, if exponent shrink slower still, we too get 0^0=0
Cas> limit(ε ^ (1/ln(-ln(ε))), ε, 0, 1)                  → 0

Let x=1/ε, we get exactly the exponent of blackpenredpen video
Cas> limit((1/x) ^ (1/ln(ln(x))), x, ∞)          → 0

Let's confirm without Cas. Let limit = a, y = ln(ln(x)), (y→∞) too
ln(a) = -e^y / y --> L'Hopital's Rule --> -e^y / 1 = -∞
a = exp(-∞) = 0      QED





We are now ready to prove blackpenredpen video limit, as x → ∞

(√(x+1) - √x) = 1/(√(x+1) + √x) ≈ (1/2) * (1/√x)      // 1 is nothing, compare against ∞

  (1/2) ^ (1/ln(ln(x))) ≈ (1/2) ^ 0 = 1
(1/√x) ^ (1/ln(ln(x))) = √((1/x) ^ (1/ln(ln(x)))) ≈ √(0) = 0

limit((√(x+1) - √x) ^ (1/ln(ln(x))), x=∞) = 1 * 0 = 0