Question

Por favor necesito ayuda con este ejercicio

Answer 1

$x-64 = \sqrt{x}^2-8^2=(\sqrt{x}-8)(\sqrt{x}+8)\\ \\ \sqrt{x}-8=\dfrac{x-64}{\sqrt{x}+8}\\ \\ ==================\\ \\ x-64 = \sqrt[3]{x}^3-4^3=(\sqrt[3]{x}-4)(\sqrt[3]{x}^2+4\sqrt[3]{x}+16)\\ \\ 4-\sqrt[3]{x}=-\dfrac{x-64}{\sqrt[3]{x}^2+4\sqrt[3]{x}+16}\\ \\ ==================\\ \\ \dfrac{\sqrt{x}-8}{4-\sqrt[3]{x}}=\dfrac{\dfrac{x-64}{\sqrt{x}+8}}{-\dfrac{x-64}{\sqrt[3]{x}^2+4\sqrt[3]{x}+16}}$

Por ende

$\dfrac{\sqrt{x}-8}{4-\sqrt[3]{x}}\sim -\dfrac{\sqrt[3]{x}^2+4\sqrt[3]{x}+16}{\sqrt{x}+8}\\ \\ \\ \lim\limits_{x\to 64}\dfrac{\sqrt{x}-8}{4-\sqrt[3]{x}}= \lim\limits_{x\to 64}-\dfrac{\sqrt[3]{x}^2+4\sqrt[3]{x}+16}{\sqrt{x}+8}\\ \\ \\ \lim\limits_{x\to 64}\dfrac{\sqrt{x}-8}{4-\sqrt[3]{x}}=\dfrac{16+16+16}{8+8}\\ \\ \\ \boxed{\lim\limits_{x\to 64}\dfrac{\sqrt{x}-8}{4-\sqrt[3]{x}}=3}$