Question

lim[1/(sinx)^2-(cosx)^2/x^2]

Answer 1

=lim[x^2-s(inx)^2(cos)^2]/[x^2(sinx)^2] （通分）
=lim[x^2-(sin2x)^2/4]/x^4（分母等价无穷小代换）
=lim[(4x^2-sin^22x]/[4x^4] （罗比达）
=lim[2x+sin2x][2x-sin2x]/(4x^4) （分子平方差）

=lim[2x+sin2x]/(4x^4) ( x->0, [2x-sin2x]等价无穷小于 4(x^3)/3 )
=lim [2x+sin2x]/3x
=4/3(罗比达)

Answer 2

=lim[1/(sinx)^2-(cosx)^2/x^2]

=lim[x^2-s(inx)^2(cos)^2[/x^2(sinx)^2]
=lim[x^2-(sin2x)^2/4]/x^4
=lim[(4x^2-sin^22x]/[4x^4]
=lim[2x+sin2x][2x-sin2x]/4x^4
=lim][2x-sin2x]/x^3
=lim][2-2cos2x]/3x^2
=2/3lim2x^2]/x^2
=4/3