Let a1,a2,....an be fixed real numbers and let
f(x)=(x−a1)(x−a2)(x−a3)...(x−an).
Find limx→a1f(x),
If a≠ a1,a2,...an,compute limx→af(x)
limx→a+1f(x)=limh→0f(a1+h)
limh→0|(a1+h−a1)(a1+h−a2)(a1+h−a3)...(a1+h−an)|
=limh→0|h(a1+h−a2)(a1+h−a3)...(a1+h−an)|
=(0×(a1−a2)×(a1−a3)×...×(a1−an)=0.
limx→a−1f(x)=limh→0f(a1−h)
=limh→0{(a1−h−a1)(a1−h−a2)(a1−h−a3)...(a1−h−an)}
=limh→0{(−h)(a1−h−a2)(a1−h−a3)...(a1−h−an)}
={0×(a1−a2)×(a1−h−a3)×...×(a1−an)}=0.
∴limx→a+1f(x)=limx→a−1f(x)=0.
Hencelimx→a1
For any a≠ a1,a2,,,,an,we have
limx→af(x)=limx→a{(x−a1)(x−a2)(x−a3)...(x−an).}
=(a−a1)(a−a2)(a−a3)...(x−an).