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

Substituting $x=4$ gives $\dfrac{0}{0}$ because of the square root, so multiplying by the conjugate removes the surd before substituting again.

Direct substitution of $x=4$ gives $\dfrac{\sqrt4-2}{4-4} = \dfrac{0}{0}$, an indeterminate form caused by the square root, so the expression is rationalised by multiplying numerator and denominator by the conjugate $\sqrt{x}+2$: $\dfrac{\sqrt{x}-2}{x-4} \times \dfrac{\sqrt{x}+2}{\sqrt{x}+2} = \dfrac{(\sqrt{x})^2 - 2^2}{(x-4)(\sqrt{x}+2)} = \dfrac{x-4}{(x-4)(\sqrt{x}+2)}$

Cancelling the common factor $(x-4)$ (valid since $x \neq 4$ throughout the limiting process): $= \dfrac{1}{\sqrt{x}+2}, \quad x \neq 4$ …

Students sometimes multiply only the numerator by the conjugate and forget to multiply the denominator by the same conjugate, which changes the value of the expression rather than merely rewriting it — the conjugate …