Question

高等数学考研

Answer 1

y'(0)=lim(t->0) [y(t)-y(0)]/t
=lim(t->0) (sint/t-1)/t
=lim(t->0) (sint-t)/t^2
=lim(t->0) (cost-1)/2t
=lim(t->0) (-t^2/2)/2t
=0
当x≠0时，y'(x)=(xcosx-sinx)/x^2
y''(0)=lim(t->0) [y'(t)-y'(0)]/t
=lim(t->0) (tcost-sint)/t^3
=lim(t->0) (-tsint)/3t^2
=-1/3
当x≠0时，y''(x)=(-xsinx-xcosx+sinx)/x^2
因为lim(x->0)y''(x)=lim(x->0)(-xsinx-xcosx+sinx)/x^2
=lim(x->0)(-sinx-xcosx+xsinx)/2x
=lim(x->0)(-2cosx+xsinx+sinx+xcosx)/2
=-1≠y''(0)
所以y''(x)在x=0点处不连续，所以当n>=3时，y^(n)(0)不存在

Answer 2

不就是1吗？