What is done first to evaluate the limit?

Take ln, so ln L = lim [ln(tan x) - ln x]/x

Apply L'Hospital directly to the power

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

Short answer

Take natural log: ln L = lim_{x→0} [ln(tan x) − ln x]/x. Using L’Hôpital’s rule or series expansion, this limit is 0, so L = e^0 = 1.

Evaluate lim_{x→0} (tan x / x)^(1/x).

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

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

Apply L'Hospital's rule:

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

Using sin 2x = 2x − (2x)³/6 + ..., we get 2/sin 2x = (1/x)(1 + 2x²/3 + ...) = 1/x + 2x/3 + ...

So ln L = lim (1/x + 2x/3 − 1/x) = 0.

(Check by series: tan x / x = 1 + x²/3 + ..., so ln(tan x / x) ≈ x²/3 and divided by x gives x/3 → 0.)

Answer: L = e⁰ = 1.

✓ Answer checked by Claude Opus 5.5 (high effort)