设f(x)=
{1-cos2xx＞0
{2x2x≤0
求f′(0)．
f′-(0)=limx→0-[f(x)-f(0)]/(x-0)=limx→0-2x2/x=0 f′+(0)=limx→0+[f(x)-f(0)]/(x-0)=limx→0+
设f(x)=
{1-cos2xx＞0
{2x2x≤0
求f′(0)．
f′-(0)=limx→0-[f(x)-f(0)]/(x-0)=limx→0-2x2/x=0 f′+(0)=limx→0+[f(x)-f(0)]/(x-0)=limx→0+
