f(x)=xarctan1/x^2（x不等于0）,f(x)=0,x=0,试讨论f'(x)在x=0处的连续性

Question

f(x)=xarctan1/x^2（x不等于0）,f(x)=0,x=0,试讨论f'(x)在x=0处的连续性

Answer 1

f(x)={xarctan(1/x^2),x≠0；
{0,x=0.
∴f'(x)=arctan(1/x^2)+x/[1+(1/x^2)^2]*(-2)/x^3
=arctan(1/x^2)-2x^2/(x^4+1),
f'(0)=lim[xarctan(1/x^2)-0]/(x-0)=π/2,
∴limf'(x)=π/2=f'(0),
∴f'(x)在x=0处连续.