求极限：1/sinx的平方-1/x的怕平方

1/(sinx)^2 - 1/x^2 = [x^2 - (sinx)^2]/[xsinx]^2
= [x^2 - (sinx)^2]/x^4*x^2/[sinx]^2
lim_{x->0}[1/(sinx)^2 - 1/x^2]
= lim_{x->0}[x^2 - (sinx)^2]/x^4*x^2/[sinx]^2
= lim_{x->0}[x^2 - (sinx)^2]/x^4
= lim_{x->0}[2x - 2sin(x)cos(x)]/[4x^3]
= lim_{x->0}[2x - sin(2x)]/[4x^3]
= lim_{x->0}[2 - 2cos(2x)]/[12x^2]
= lim_{x->0}[1 - cos(2x)]/[6x^2]
= lim_{x->0}[2sin(2x)]/[12x]
= 2/6lim_{x->0}[sin(2x)]/[2x]
= 2/6
= 1/3