Q.$\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}$

The key idea is that direct substitution gives the indeterminate form $\frac{0}{0}$, so we factor both numerator and denominator to cancel the common factor.

Factor the numerator: $3x^2 - x - 10 = (3x + 5)(x - 2)$.

Factor the denominator: $x^2 - 4 = (x - 2)(x + 2)$.

Cancel the common factor $(x - 2)$ (valid since $x \neq 2$ in the limit):

Why this approach works

When you plug $x = 2$ into the expression $\dfrac{3x^2 - x - 10}{x^2 - 4}$, you get $\dfrac{3(4) - 2 - 10}{4 - 4} = \dfrac{0}{0}$. That's an indeterminate form — it tells us nothing about the actual limit. The function might approach a finite number, go to infinity, or oscillate.

The key insight: both numerator and denominator are polynomials. A limit of a polynomial quotient that gives $\frac{0}{0}$ means $x=2$ is a root of both polynomials. So $(x-2)$ is a common factor. Factor them, cancel the factor that causes the trouble, and the remaining expression is continuous at $x=2$ — then you can just substitute.

Step-by-step solution

1. Factor the denominator.

The denominator is $x^2 - 4$, a difference of squares:

2. Factor the numerator.

We need two numbers whose product is $3 \times (-10) = -30$ and whose sum is $-1$ (the coefficient of $x$). Those numbers are $-6$ and $5$. So:

$3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x-2) + 5(x-2) = (x-2)(3x+5).$ …