## Calculus: Early Transcendentals (2nd Edition)

We have to determine $L=\lim\limits_{x \to 4} \dfrac{3(x-4)\sqrt{x+5}}{3-\sqrt{x+5}}$. Multiply both numerator and denominator by the conjugate of the denominator: $L=\lim\limits_{x \to 4} \dfrac{3(x-4)\sqrt{x+5}}{3-\sqrt{x+5}}\cdot\dfrac{3+\sqrt{x+5}}{3+\sqrt{x+5}}$ $=\lim\limits_{x \to 4} \dfrac{3(x-4)\sqrt{x+5}(3+\sqrt{x+5})}{3^2-(\sqrt{x+5})^2}$ $=\lim\limits_{x \to 4} \dfrac{3(x-4)\sqrt{x+5}(3+\sqrt{x+5})}{9-(x+5)}$ $=\lim\limits_{x \to 4} \dfrac{3(x-4)\sqrt{x+5}(3+\sqrt{x+5})}{4-x}$ Simplify: $L=-\lim\limits_{x \to 4} 3\sqrt{x+5}(3+\sqrt{x+5})$ Determine the limit: $L=-3\sqrt{4+5}(3+\sqrt{4+5})=-3(3)(3+3)=-54$