Q.Find the value of $k$ so that $f(x) = \dfrac{\sqrt{x} - 2}{x - 4}$ (for $x \neq 4$), $f(4) = k$, is continuous at $x = 4$.

The limit is a $\tfrac00$ surd form, so rationalise the numerator before evaluating; then set $k$ equal to the limit.

Continuity at $x = 4$ requires $k = f(4) = \displaystyle\lim_{x\to 4} f(x)$. Direct substitution gives $\dfrac{\sqrt4 - 2}{4-4} = \dfrac{0}{0}$, an indeterminate form, so rationalise the numerator.

Rationalise. Multiply numerator and denominator by the conjugate $\sqrt{x} + 2$:

For $x \neq 4$ the factor $x - 4$ cancels, leaving $\dfrac{1}{\sqrt{x}+2}$.

Take the limit. $\displaystyle\lim_{x\to 4} f(x) = \lim_{x\to 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt4 + 2} = \frac{1}{2 + 2} = \frac{1}{4}$.

Set the constant. For continuity, $k = \displaystyle\lim_{x\to 4} f(x) = \dfrac{1}{4}$. …

Alternative — substitution $t=\sqrt{x}$: then $x=t^2$ and $x-4=t^2-4=(t-2)(t+2)$, so $\dfrac{\sqrt{x}-2}{x-4}=\dfrac{t-2}{(t-2)(t+2)}=\dfrac{1}{t+2}$. As $ …

A common error is to attempt to cancel $\sqrt{x}-2$ against $x-4$ directly without rationalising — but $x-4$ is not written in terms of $\sqrt{x}$, so no cancellation is visible. Rationalising with the conjugate (or the substitution $t= …