limit x tends to 0 (1/x^2 - 1/(sinx)^2 )

Dear tushar

Lt_x→0(1/x^2 - 1/(sinx)^2 )

or

Lt_x→0(sin²x-x²)/x²sin²x

or Lt_x→0(sin²x-x²)/x⁴{sin²x/x²}

or Lt_x→0(sin²x-x²)/x⁴     because  Lt_x→0 sinx/x     ⁼¹

Use L hosipal rule

 limit will be =-1/3