Evaluating a 0/0 limit by rationalizing a radical expression

Direct substitution at $x=0$ gives $\frac{2-\sqrt{4}}{0} = \frac{0}{0}$, an indeterminate form, so we need an algebraic technique. Multiply numerator and denominator by the conjugate of the numerator, $2+\sqrt{x+4}$: $\frac{2-\sqrt{x+4}}{x}\cdot\frac{2+\sqrt{x+4}}{2+\sqrt{x+4}} = \frac{4-(x+4)}{x\left(2+\sqrt{x+4}\right)}.$ The numerator simplifies to $4-(x+4) = -x$.

This leaves $\frac{-x}{x\left(2+\sqrt{x+4}\right)} = \frac{-1}{2+\sqrt{x+4}}$ after canceling the common factor of $x$ (valid since $x \to 0$ means $x \neq 0$). Now substitute $x=0$ directly: $\frac{-1}{2+\sqrt{4}} = \frac{-1}{4}.$

The conjugate used was two plus the square root of x plus four. This matches the numerator's structure, two minus the square root of x plus four, but with the opposite sign between the terms. Multiplying by this conjugate over itself eliminates the square root using the difference of squares pattern, turning the numerator into a simple expression that cancels with x in the denominator.