利用級(jí)數(shù)可以做吧,
tanx=x+x^3/3+2x^5/15+O(x^6)=T+O(x^6),
tanT=T+T^3/3+2T^5/15+O(T^6)=x+2x^3/3+3x^5/5+O(x^6);
sinx=x-x^3/6+x^5/120+O(x^6)=S+O(x^6),
sinS=S-S^3/6+S^5/120+O(S^6)=x-x^3/3+x^5/10+O(x^6).
則
lim[tan(tanx)-sin(sinx)]/x^3,x->0
=lim[(x+2x^3/3+3x^5/5)-(x-x^3/3+x^5/10)]/x^3,x->0
=lim(x^3+x^5/2)/x^3,x->0
=1.