求极限lim_x→0[tan(2x+x²)/arcsin3x]．

求极限lim_x→0[tan(2x+x²)/arcsin3x]．

令t=arcsin3x，则x→0⟺t→0 因此lim_x→0(arcin3x/3x)=lim_t→0=1 所以当x→0时，arcsin3x～3x 于是lim_x→0[tan(2x+x²)/arcsin3x]=lim_x→0[tan(2x+x²)/3x] =lim_x→0[tan(2x+x²)/(2x+x²)]•[(2x+x²)/3x] =lim_x→0[sin(2x+x²)/(2x+x²)]•[1/cos(2x+x²)•(2+x)/3=2/3．