x≠0时，f'(x)=[(e^x-cosx)/x]'=[(e^x+sinx)x-(e^x-cos)]/x^2=(xe^x-e^x+xsinx+cosx)/x^2
x->0时，lim f(x)=lim(e^x-cosx)'/x'=lim (e^x+sinx)=1, 故f(x)在x=0处连续；
故有 f'(0)=lim(d->0) [f(d)-f(0)]/d=lim[(e^d-cosd)/d-1]/d=lim(e^d-cosd-d)/d^2=lim(e^d+sind-1)/(2d)
=lim(e^d+cosd)/2=(1+1)/2=1
x->0时，lim f(x)=lim(e^x-cosx)'/x'=lim (e^x+sinx)=1, 故f(x)在x=0处连续；
故有 f'(0)=lim(d->0) [f(d)-f(0)]/d=lim[(e^d-cosd)/d-1]/d=lim(e^d-cosd-d)/d^2=lim(e^d+sind-1)/(2d)
=lim(e^d+cosd)/2=(1+1)/2=1
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