求极限~lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)

lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1+1-cosx]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1+2sin^2(x/2)]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1]/sin^2(x/2)+2 （[(1+x)^1/n]-1~（1/n）*x ）
=lim(x→0)1/2xsinx/sin^2(x/2)+2
=lim1/2x^2/ (x/2)^2+2
=2+2
=4