Question

lim（x→0）[1/（sinx）^2-（cosx）^2/x^2]

Answer 1

lim（x→0）[1/（sinx）^2-（cosx）^2/x^2]
=lim（x→0）[x^2-sin^2 x（cosx）^2]/[x^2（sinx）^2]
=lim（x→0）[x^2-sin^2 x（cosx）^2]/[x^4]
=lim（x→0）[x^2-1/4sin^2 (2x)]/[x^4] (0/0)
=lim（x→0）[2x-sin (2x)cos(2x)]/[4x^3]
=lim（x→0）[2x-1/2sin (4x)]/[4x^3] (0/0)
=lim（x→0）[2-2cos (4x)]/[12x^2]
=lim（x→0）(4x)^2/[12x^2]
=4/3

Answer 2

lim(x->0)1/[1/sinx^2-cosx^2/x^2
=lim(x->0)x^2sinx^2/(x^2-sinx^2cosx^2)
=lim(x->0)x^2sinx^2/(x^2-(1/4)(sin2x)^2
=lim(x->0)(2xsinx^2+x^2sin2x)/(2x-(1/2)sin4x)
=lim(x->0)(2sinx^2+4xsin2x+2x^2cos2x)/(2-2cos4x)
=lim(x->0)(6sin2x+12xcos2x-4x^2sin2x)/(8sin4x)
=lim(x->0) (24cos2x-32xsin2x-8x^2cos2x)/(32cos4x)
=24/32=3/4