Let $a_{1}, a_{2}, \ldots ., a_{n}$ be fixed real numbers and define a function
$f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)$
What is $\lim _{x \rightarrow a} f(x)$ ? For some $a \neq a_{1}, a_{2} \ldots, a_{n,}$ compute $\lim _{x \rightarrow a} f(x)$.
The given function is $f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)$
$\lim _{x \rightarrow a_{1}} f(x)=\lim _{x \rightarrow a_{1}}\left[\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)\right]$
$=\left[\lim _{x \rightarrow a_{1}}\left(x-a_{1}\right)\right]\left[\lim _{x \rightarrow a_{1}}\left(x-a_{2}\right)\right] \ldots\left[\lim _{x \rightarrow a_{1}}\left(x-a_{n}\right)\right]$
$=\left(a_{1}-a_{1}\right)\left(a_{1}-a_{2}\right) \ldots\left(a_{1}-a_{n}\right)=0$
$\therefore \lim _{x \rightarrow 0_{1}} f(x)=0$
Now, $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}\left[\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)\right]$
$=\left[\lim _{x \rightarrow a}\left(x-a_{1}\right)\right]\left[\lim _{x \rightarrow a}\left(x-a_{2}\right)\right] \ldots\left[\lim _{x \rightarrow a}\left(x-a_{n}\right)\right]$
$=\left(a-a_{1}\right)\left(a-a_{2}\right) \ldots . .\left(a-a_{n}\right)$
$\therefore \lim _{x \rightarrow a} f(x)=\left(a-a_{1}\right)\left(a-a_{2}\right) \ldots\left(a-a_{n}\right)$
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