函数f(x)在x=0点连续,且极限lim(f(x)+3)/x=2,问函数f(x)在x=0点是否可导

lim(x->0) (f(x)+3)/x=2
∴ x->0 时,(f(x)+3) = O(x) 即:
lim(x->0) (f(x)+3) = 0 ,又函数f(x)在x=0点连续:
∴ lim(x->0) f(x) = -3 = f(0)
2=lim(x->0) (f(x)+3)/x
=lim(x->0) [f(x)-f(0)]/(x-0)
= f'(0)
即：函数f(x)在x=0点可导 ；f'(0) = 2.