f(x)={g(x)sin1/x x≠0
{0,x=0
f'(0)=lim {x趋向0}([g(x)sin1/x-0)/x=lim[g(x)sin1/x]/x =lim g'(x)sin1/x+g(x)dsin1/x/dx=lim g'(x)sin1/x+g(x)cos1/x*(-1/x^2)
因为x趋向0时,sin1/x,cos1/x都是不定值
故g'(0)=0,g(0)=0时,f(x)在x=0点导数存在,即f(x)在0点可微