• Matéria: Matemática
  • Autor: juniormendes89
  • Perguntado 9 anos atrás

qual é o lim     \sqrt[4]{x} - \sqrt[4]{2}

               x->2                    x-2

Respostas

respondido por: Celio
64

Olá, Jr.

 

<var>\lim\limits_{x\to2}\frac{\sqrt[4]{x} - \sqrt[4]{2}}{x-2}=\lim\limits_{x\to2}\frac{\sqrt[4]{x} - \sqrt[4]{2}}{(\sqrt{x}+\sqrt2)(\sqrt{x}-\sqrt2)}=\\\\=\lim\limits_{x\to2}\frac{\sqrt[4]{x} - \sqrt[4]{2}}{(\sqrt{x}+\sqrt2)(\sqrt[4]{x}+\sqrt[4]2)(\sqrt[4]{x}-\sqrt[4]2)}=\lim\limits_{x\to2}\frac{1}{(\sqrt{x}+\sqrt2)(\sqrt[4]{x}+\sqrt[4]2)}=\\\\=\frac{1}{(\sqrt{2}+\sqrt2)(\sqrt[4]{2}+\sqrt[4]2)}=\frac{1}{2\sqrt{2} \cdot 2\sqrt[4]2}=\frac{1}{4\sqrt[4]{2^2} \cdot \sqrt[4]2}=\frac{1}{4\cdot\sqrt[4]{2^3}}=</var>

 

<var>=\frac{\sqrt[4]{2}}{4\cdot\sqrt[4]{2^3}\cdot\sqrt[4]2}}=\frac{\sqrt[4]{2}}{4\cdot\sqrt[4]{2^3\cdot2}}=\frac{\sqrt[4]{2}}{4\cdot\sqrt[4]{2^4}}=\frac{\sqrt[4]{2}}{4\cdot2}=\frac{\sqrt[4]{2}}{8}</var>

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