Limit of (2−x/a)^(tan(πx/2a))

Short answer

This is also form 1^∞. Take ln, apply L'Hospital: ln L = lim ln(2 - x/a)/cot(pi x/(2a)) = 2/pi. Thus L = e^(2/pi).

Evaluate lim_{x→a} (2 − x/a)^(tan(πx/2a)).

Let L = lim_{x→a} (2 − x/a)^(tan(πx/2a)). At x = a the base → 1 and the exponent → ∞: form 1^∞.

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

Apply L'Hospital's rule. Derivative of the numerator: (−1/a)/(2 − x/a). Derivative of the denominator: −(π/2a) cosec²(πx/2a).

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

Answer: L = e^(2/π). (Putting x = at reduces this to the previous problem, lim_{t→1} (2 − t)^(tan(πt/2)).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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