Limit of (1/x²)^(tan x)

Short answer

Limit is of the form ∞^0. Take logarithm: ln L = -2 lim tan x ln|x|. Apply L'Hospital's rule to get ln L = 0, so L = e^0 = 1.

Evaluate lim_{x→0} (1/x²)^(tan x).

Let L = lim_{x→0} (1/x²)^(tan x). As x → 0, 1/x² → ∞ and tan x → 0, so this is the form ∞⁰.

Take logarithms:

ln L = lim tan x · ln(1/x²) = −2 lim tan x · ln|x|

First let x → 0⁺ (so |x| = x):

ln L = −2 lim (ln x)/(cot x) (form ∞/∞)

Apply L'Hospital's rule:

ln L = −2 lim (1/x)/(−cosec²x)
     = 2 lim sin²x / x
     = 2 lim (sin x/x) · sin x
     = 2 · 1 · 0 = 0

As x → 0⁻, tan x · ln(1/x²) only changes sign (tan x is odd and ln(1/x²) is even), so it also tends to 0. Hence ln L = 0 from both sides.

Answer: L = e⁰ = 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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