Q.Let $a_1, a_2, \ldots, a_n$ be fixed real numbers and define a function $f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$. What is $\lim_{x\to a_1} f(x)$? For some $a \neq a_1, a_2, \ldots, a_n$, compute $\lim_{x\to a} f(x)$.

Concept: Limit of a Polynomial — a polynomial is continuous everywhere, so the limit at any point is simply the function value at that point.

Since $f(x)$ is a polynomial (a product of linear factors), it is continuous for all real $x$.

For $x \to a_1$:

For $x \to a$ where $a \neq a_1, a_2, \ldots, a_n$:

$\lim_{x\to a} f(x) = f(a) = (a - a_1)(a - a_2)\cdots(a - a_n),$ …

The function $f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$ is a polynomial — specifically, a product of $n$ linear factors. Polynomials are continuous everywhere on the real line. That’s the core idea: continuity means the limit as $x$ approaches any point is just the function’s value at that point.

So the problem reduces to: what is $f(a_1)$? And what is $f(a)$ for some $a$ that isn’t any of the $a_i$?

1. At $x = a_1$:

   The factor $(x - a_1)$ becomes $0$. Multiplying by anything else gives $0$. So $f(a_1) = 0$.

   By continuity, $\displaystyle \lim_{x \to a_1} f(x) = f(a_1) = 0$.

2. At $x = a$ (where $a \neq a_1, a_2, \ldots, a_n$):

   None of the factors $(a - a_i)$ is zero, so $f(a)$ is a finite real number — the product of $n$ non-zero differences.

   Again by continuity, $\displaystyle \lim_{x \to a} f(x) = f(a) = (a - a_1)(a - a_2)\cdots(a - a_n)$. …