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# Let $a_1, a_2, ....,a_n$ be fixed real numbers and define a function $f(x) = (x-a_1)(x-a_2)...(x-a_n)$  What is $\lim\limits_{x \to a_1}f(x)$? For some $a \neq a_1, a_2,...,a_n$. Compute $\lim\limits_{x \to a}f(x)$.

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The given function is $f(x) = (x-a_1)(x-a_2)...(x-a_n)$
$\lim\limits_{x \to a_1} f(x) = \lim\limits_{x \to a_1} [(x-a_1)(x-a_2)...(x-a_n)]$
$= [ \lim\limits_{x \to a_1} (x-a_1)] [ \lim\limits_{x \to a_1} (x-a_2)]...[\lim\limits_{x \to a_1} (x-a_n)]$
$= (a_1-a_1)(a_1-a_2)...(a_1-a_n)=0$
$\therefore \lim\limits_{x \to a_1} f(x)=0$
Now $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} [(x-a_1)(x-a_2)...(x-a_n)]$
$= [ \lim\limits_{x \to a} (x-a_1)] [ \lim\limits_{x \to a} (x-a_2)]...[\lim\limits_{x \to a} (x-a_n)]$
$= (a-a_1)(a-a_2)...(a-a_n)$
$\therefore \lim\limits_{x \to a} f(x) = (a-a_1)(a-a_2)...(a-a_n)$
answered Apr 7, 2014