ln L is rewritten as:

ln(2-x)/cot(pi x/2) of type 0/0

Limit of (2−x)^(tan(πx/2))

Short answer

Form 1^∞. Take ln: ln L = lim tan(pi x/2) ln(2-x) = lim ln(2-x)/cot(pi x/2). Apply L'Hospital to get ln L = 2/pi, so L = e^(2/pi).

Evaluate lim_{x→1} (2−x)^(tan(πx/2)).

Let L = lim_{x→1} (2 − x)^(tan(πx/2)). At x = 1: 2 − x → 1 and tan(πx/2) → ∞, so this is the form 1^∞.

Take logarithms:

ln L = lim tan(πx/2) · ln(2 − x) = lim ln(2 − x) / cot(πx/2) (form 0/0)

Apply L'Hospital's rule:

ln L = lim [−1/(2−x)] / [−(π/2) cosec²(πx/2)]
     = lim (2/π) · sin²(πx/2) / (2 − x)
     = (2/π) · 1/1 = 2/π

Answer: L = e^(2/π).

✓ Answer checked by Claude Opus 5.5 (high effort)