可导性是左右导数相同,连续性是左右极限相同
f(x)
=[√(1+x) -1]/√x ; x>0
=0 ; x≤0
f(0+)
=lim(x->0+) [√(1+x) -1]/√x
=lim(x->0+) (1/2)x/√x
=0
f(0)=f(0-) =0
x=0 , f(x) 连续
f'(0+)
= lim(h->0){ [√(1+h) -1]/√h - f(0) } /h
= lim(h->0){ [√(1+h) -1]/√h - 0 } /h
= lim(h->0) [√(1+h) -1] /h^(3/2)
= lim(h->0) (1/2)h /h^(3/2)
= lim(h->0) (1/2)/h^(1/2)
不存在
=> x=1, f(x) 不可导