Question:

Let a1, a2,..., an be fixed real numbers and define a function f(x)=(x-a1)(x-a2)...(x-an).
What is limxa1\lim_{x\rightarrow a_1} f(x)? 
For some a ≠ a1, a2 .....an, Compute limxa\lim_{x\rightarrow a} f(x).

Updated On: Oct 25, 2023
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Solution and Explanation

The given function is f(x) = (x − a1) ( x − a2)... ( x − an)
limxa1\lim_{x\rightarrow a_1} f(x)= limxa1\lim_{x\rightarrow a_1} [(x-a1)(x − a2).....(x − an)]
=[limxa1\lim_{x\rightarrow a_1} (x − a1) ][limxa1\lim_{x\rightarrow a_1}(x − a2)]..... [limxa1\lim_{x\rightarrow a_1} (x-an)]
=(a1-a1)(a1-a2)....(a1 -an) = 0
limxa1\lim_{x\rightarrow a_1} f(x)=0
Now, limxa\lim_{x\rightarrow a} f(x)= limxa\lim_{x\rightarrow a}[(x − a1)(x-a2)...(x-an)]
= [limx(xa1)\lim_{x\rightarrow (x-a_1)}][limxa\lim_{x\rightarrow a} (x-a2)]...limxa\lim_{x\rightarrow a}(x-an)]
=(a-a1) (a-a2)....(a-an)
limxa\lim_{x\rightarrow a} f(x)=(a-a1)(a-a2)...(a-an)
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