lim(x->0)[√(1+tanx)- √(sinx+1) ]/x^3
=(1/2)lim(x->0)(tanx - sinx)/(x^3) (0/0)
=(1/2)lim(x->0)((secx)^2 - cosx)/(3x^2) (0/0)
=(1/2)lim(x->0)(2(secx)^2.tanx + sinx)/(6x) (0/0)
=(1/2)lim(x->0){ 2[(secx)^4+ 2(tanx)^2.(secx)^2] + cosx }/6
=[ 2(1+ 0) + 1 ]/12
=1/4