顯然x趨于0的時候,1-cos2x趨于0,
那么√[1+f(x)(1-cos2x)] - 1也趨于0,
而且等價于0.5* f(x)(1-cos2x),
而1-cos2x等價于0.5*(2x)^2即2x^2,
所以√[1+f(x)(1-cos2x)] - 1等價于 x^2 *f(x),
而x趨于0時,sinx也是x的等價無窮小,
故
lim(x趨于0) {√[1+f(x)(1-cos2x)] - 1} /xsinx
=lim(x趨于0)x^2 *f(x) /x^2
=lim(x趨于0)f(x)
=6