Let a1,a2,⋅⋅⋅⋅,an be fixed real numbers and define a function f(x)=(x−a1)(x−a2).....(x−an) What is limx→a1f(x)? For some a≠a1,a2⋅⋅⋅,an, compute limx→af(x)⋅
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Solution
We have f(x)=(x−a1)(x−a2).....(x−an) limx→a1f(x)=limx→a3[(x−a1)(x−a2)⋅⋅⋅(x−an)] = [limx→a1(x−a1)][limx→a1(x−a2)]...[limx→a1(x−an)] = (a1−a)(a1−a2)...(a1−an)=0 ∴limx→a1f(x)=0 Now limx→af(x)=limx→a[(x−a1)(x−a2)⋅⋅⋅(x−an)] = [limx→a(x−a1)][limx→a(x−a2)]...[limx→a(x−an)] = (a−a1)(a−a2)...(a−an) ∴limx→af(x)=(a−a1)(a−a2)...(a−an)