Limit of (sin x / x)^(1/x)

Let L = lim_{x→0⁺} (sin x / x)^(1/x). As x → 0, sin x / x → 1 and 1/x → ∞, so this is the form 1^∞.

Take logarithms:

ln L = lim_{x→0⁺} ln(sin x / x) / x = lim [ln sin x − ln x] / x (form 0/0)

Apply L'Hospital's rule:

ln L = lim [ cot x − 1/x ]
     = lim (x cos x − sin x) / (x sin x)      (0/0)
     = lim (−x sin x) / (sin x + x cos x)     (differentiate top and bottom)
     = 0 / 0 → differentiate again:
       lim (−sin x − x cos x) / (2 cos x − x sin x) = 0 / 2 = 0

Check by series: sin x / x = 1 − x²/6 + ..., so ln(sin x/x) ≈ −x²/6, and (−x²/6)/x = −x/6 → 0.

So ln L = 0.

Answer: L = e⁰ = 1.