Limit of 1/x² − 1/sin²x

The limit is of the form ∞ − ∞. Combine the fractions:

1/x² − 1/sin²x = (sin²x − x²) / (x² sin²x) (form 0/0)

Use the series sin x = x − x³/6 + x⁵/120 − ...

sin²x = x² − x⁴/3 + 2x⁶/45 − ...
sin²x − x² = −x⁴/3 + 2x⁶/45 − ...
x² sin²x   = x⁴ − x⁶/3 + ...

Divide numerator and denominator by x⁴:

(−1/3 + 2x²/45 − ...) / (1 − x²/3 + ...) → −1/3 as x → 0.

(Same result by L'Hospital's rule applied four times, since both numerator and denominator vanish to order x⁴.)

Answer: −1/3.