Q.Let $f(x) = \begin{cases} x^2 - 1, & 0 < x < 2 \\ 2x + 3, & 2 \leq x < 3 \end{cases}$, the quadratic equation whose roots are $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$ is
(A) $x^2 - 6x + 9 = 0$
(B) $x^2 - 7x + 8 = 0$
(C) $x^2 - 14x + 49 = 0$
(D) $x^2 - 10x + 21 = 0$

Concept: Limit Of Polynomial — for a polynomial (or any function continuous at the point), the limit is simply the value of the function at that point. Here we evaluate one-sided limits using the appropriate piece.

Step 1: For $x \to 2^-$, we use $f(x) = x^2 - 1$ (since $0 < x < 2$).

$\lim_{x \to 2^-} f(x) = 2^2 - 1 = 3$.

Step 2: For $x \to 2^+$, we use $f(x) = 2x + 3$ (since $2 \leq x < 3$). …

The core idea here is that a piecewise function can have different behaviours from the left and right of a point. The limits at that point are found by plugging into the appropriate piece — no need to worry about the function’s actual value at $x=2$ (which isn’t even defined in the given domain). Once you have the two numbers, constructing a quadratic with them as roots is straightforward.

Let’s work through it step by step.

1. Identify the left-hand limit $\lim_{x \to 2^-} f(x)$. For $x$ approaching $2$ from the left, we use the piece $f(x) = x^2 - 1$, valid for $0 < x < 2$. Since $x^2 - 1$ is a polynomial, it is continuous everywhere, so the limit is simply the value at $x=2$:

2. Identify the right-hand limit $\lim_{x \to 2^+} f(x)$. For $x$ approaching $2$ from the right, we use $f(x) = 2x + 3$, valid for $2 \leq x < 3$. Again, this is a polynomial (linear) and continuous, so: $\lim_{x \to 2^+} f(x) = 2(2) + 3 = 4 + 3 = 7.$ …