f(x)在x=2处连续,lim(x→2)f(x)/(x-2)=3,
因为(x-2)->0,所以f(x)含有因式(x-2),即f(x)=(x-2)g(x),
从而f(2)=0,所以
f'(2)=lim(x→2)[f(x)-f(2)]/(x-2)
=lim(x→2)f(x)/(x-2)
=3
f(x)在x=2处连续,lim(x→2)f(x)/(x-2)=3,
因为(x-2)->0,所以f(x)含有因式(x-2),即f(x)=(x-2)g(x),
从而f(2)=0,所以
f'(2)=lim(x→2)[f(x)-f(2)]/(x-2)
=lim(x→2)f(x)/(x-2)
=3