![\lim_{x\rightarrow - \infty}(\sqrt[3]{x^7}+ x)/\sqrt[3]{x^2}- 1 \lim_{x\rightarrow - \infty}(\sqrt[3]{x^7}+ x)/\sqrt[3]{x^2}- 1](/latexrender/pictures/8c36bbe16b753894dec8c05751247414.png)
socorro? kkkk
Tentei transformar as raizes em potencias e resolver pela regra do polonimio, mas nao deu certo... o que devo fazer?
![\lim_{x\rightarrow - \infty}(\sqrt[3]{x^7}+ x)/\sqrt[3]{x^2}- 1 \lim_{x\rightarrow - \infty}(\sqrt[3]{x^7}+ x)/\sqrt[3]{x^2}- 1](/latexrender/pictures/8c36bbe16b753894dec8c05751247414.png)



...
![\lim_{x \to - \infty} \frac{\sqrt[3]{x^7} + x}{\sqrt[3]{x^2} +1} = \lim_{x \to - \infty} \frac{\sqrt[3]{x^7}}{\sqrt[3]{x^2}} \cdot \frac{(1 + x^{\frac{-4}{7}})}{(1 + x^{\frac{-2}{3}})} = \lim_{x \to - \infty} \sqrt[3]{x^5} \cdot \frac{(1 + x^{\frac{-4}{7}})}{(1 + x^{\frac{-2}{3}})} \lim_{x \to - \infty} \frac{\sqrt[3]{x^7} + x}{\sqrt[3]{x^2} +1} = \lim_{x \to - \infty} \frac{\sqrt[3]{x^7}}{\sqrt[3]{x^2}} \cdot \frac{(1 + x^{\frac{-4}{7}})}{(1 + x^{\frac{-2}{3}})} = \lim_{x \to - \infty} \sqrt[3]{x^5} \cdot \frac{(1 + x^{\frac{-4}{7}})}{(1 + x^{\frac{-2}{3}})}](/latexrender/pictures/d81be51a615fdfeec6b7e4ff225bd7c7.png)


??
( raiz do maior coeficiente) deu 1...
Voltar para Cálculo: Limites, Derivadas e Integrais
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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}} \frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png)
(dica : igualar a expressão a
e elevar ao quadrado os dois lados)