Limit of (tan x / x)^(1/x²)

Short answer

Let L = lim_{x→0} (tan x/x)^{1/x²}. Then ln L = lim ln(tan x/x)/x² = 1/3 (by series or L'Hôpital). Hence L = e^{1/3}.

Evaluate lim_{x→0} (tan x / x)^(1/x²).

Let L = lim_{x→0} (tan x / x)^(1/x²). This is the form 1^∞.

ln L = lim_{x→0} ln(tan x / x) / x² = lim [ln tan x − ln x] / x² (form 0/0)

Apply L'Hospital's rule:

ln L = lim [ sec²x/tan x − 1/x ] / (2x)
     = lim [ 2/sin 2x − 1/x ] / (2x)
     = lim (2x − sin 2x) / (2x² sin 2x)

Use series sin 2x = 2x − (2x)³/6 + ... = 2x − 4x³/3 + ...

Numerator: 2x − sin 2x = 4x³/3 + ... Denominator: 2x² · 2x = 4x³ + ...

ln L = (4/3)/4 = 1/3.

(Direct series check: tan x / x = 1 + x²/3 + ..., so ln(tan x / x) = x²/3 + ..., and dividing by x² gives 1/3.)

Answer: L = e^(1/3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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