f'(0) 存在,且lim(x趋向于0) 1/x[f(x)-f(x/3)]=a,求'f(0)
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lim(x->0)[f(x)-f(x/3)]/x
=lim(x->0)[f(x)-f(0)+f(0)-f(x/3)]/x
=lim(x->0)[f(x)-f(0)]/(x-0)-lim(x->0)[f(x/3)-f(0)]/x
=f'(0)-lim(x->0)(1/3)*[f(x/3)-f(0)]/(x/3-0)
=f'(0)-1/3*lim(x/3->0)[f(x/3)-f(0)]/(x/3-0)
=f'(0)-1/3*f'(0)
=2/3*f'(0)
即2/3*f'(0)=a
所以f'(0)=3a/2
=lim(x->0)[f(x)-f(0)+f(0)-f(x/3)]/x
=lim(x->0)[f(x)-f(0)]/(x-0)-lim(x->0)[f(x/3)-f(0)]/x
=f'(0)-lim(x->0)(1/3)*[f(x/3)-f(0)]/(x/3-0)
=f'(0)-1/3*lim(x/3->0)[f(x/3)-f(0)]/(x/3-0)
=f'(0)-1/3*f'(0)
=2/3*f'(0)
即2/3*f'(0)=a
所以f'(0)=3a/2
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