The value of _ x 1 x¹⁵-1 x¹⁰-1 =

Step-by-step solution

The limit is of the form 0 0 as x 1 . Using L'Hopital's rule, we differentiate the numerator and the denominator with respect to x : _ x 1 d dx (x¹⁵ - 1) d dx (x¹⁰ - 1) = _ x 1 15x¹⁴ 10x^9 Simplifying the expression: _ x 1 15 10 x¹⁴⁻⁹ = 3 2 _ x 1 x^5 Substituting x = 1 : 3 2 (1)^5 = 3 2 Answer: 3 2