Q.Evaluate $\lim_{x \to 2} \dfrac{x^2 - 4}{\sqrt{3x - 2} - \sqrt{x + 2}}$.

Concept: Limit Of Polynomial (with rationalisation)

The denominator becomes $0$ at $x=2$, so we rationalise it.

Step 1 – Multiply numerator and denominator by the conjugate

Step 2 – Simplify denominator

Step 3 – Factor numerator and cancel

So the expression becomes

The core idea here is that when a limit gives $\frac{0}{0}$, we need to algebraically rewrite the expression so the cancellation becomes visible. The denominator has square roots, and a classic trick is to rationalise — multiply top and bottom by the conjugate of the denominator. This removes the square roots and often reveals a common factor.

Let’s walk through it.

1. Check the form

   Substitute $x = 2$:

   Numerator: $2^2 - 4 = 0$

   Denominator: $\sqrt{3(2)-2} - \sqrt{2+2} = \sqrt{4} - \sqrt{4} = 0$

   So it’s $\frac{0}{0}$ — indeterminate. We must simplify.

2. Rationalise the denominator

   The conjugate of $\sqrt{3x-2} - \sqrt{x+2}$ is $\sqrt{3x-2} + \sqrt{x+2}$. Multiply numerator and denominator by this conjugate:

3. Simplify the denominator Using $(a-b)(a+b) = a^2 - b^2$:

So the expression becomes:

4. Factor everything Notice $x^2 - 4 = (x-2)(x+2)$ and $2x - 4 = 2(x-2)$. So:

The $(x-2)$ cancels (for $x \neq 2$, which is fine in a limit):