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Multiplying by the conjugate resolves the limit by eliminating the radical in the denominator, which causes the $0/0$ indeterminacy. For $\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}$, multiply the numerator and denominator by $\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}$.

Conditions: The limit is as $x \to 5$.; The expression involves a radical in the denominator causing a $0/0$ form.; The conjugate is $\sqrt{x+4}+3$.

Multiplying by the conjugate resolves the limit by eliminating the radical in the denominator, which causes the $0/0$ indeterminacy. For $\lim_{x\to5}\frac{x-5}{\sqrt{x+4}-3}$, multiply the numerator and denominator by $\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}$.

Conditions: The limit is as $x \to 5$.; The expression involves a radical in the denominator causing a $0/0$ form.; The conjugate is $\sqrt{x+4}+3$.