讨论函数f(x)={x^2sin1/x and 0在x=0处的连续性与可导性

lim(x-->0)x^2sin(1/x)
=lim(x-->0)x*sin(1/x)/(1/x)
=0
即lim(x->0)=f(0)
所以f(x)在x=0处连续.
lim(Δx-->0)[f(0+Δx)-f(0)]/Δx
lim(Δx-->0)Δxsin(1/Δx)
=lim(Δx-->0)sin(1/Δx)/(1/Δx)
=1
f(x)在x=0处可导.