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Question Number 170766 by cortano1 last updated on 30/May/22

$$\underset{x\to -\mathrm{\infty }}{\lim }\sqrt[7]{x^7+x^6-1}+\sqrt[9]{x^5+1-x^9}=?$$

Answered by aleks041103 last updated on 30/May/22

$$\sqrt[7]{x^7+x^6-1}=x\sqrt[7]{1+\frac{1}{x}-\frac{1}{x^7}}\sim x(1+\frac{x^6-1}{7x^7})$$$$\sqrt[9]{-x^9+x^5+1}=-x\sqrt[9]{1-(\frac{1}{x^4}+\frac{1}{x^9})}\sim x(1-\frac{x^5+1}{9x^9})$$$$\Rightarrow L=\underset{x\to -\mathrm{\infty }}{lim}\frac{x^6-1}{7x^6}-\frac{x^5+1}{9x^8}=\frac{1}{7}$$

Answered by greougoury555 last updated on 31/May/22

$$=\underset{x\to 0}{\lim }\sqrt[7]{\frac{1}{x^7}+\frac{1}{x^6}-1}+\sqrt[9]{\frac{1}{x^5}-\frac{1}{x^9}+1}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt[7]{1+x-x^7}+\sqrt[9]{x^4-1+x^9}}{x}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{1+x-x^7}-\sqrt[9]{1-(x^4+x^9)}}{x}$$$$=\underset{x\to 0}{\lim }\frac{1+\left(\frac{x-x^7}{7}\right)-(1-\left(\frac{x^4+x^9}{9}\right))}{x}$$$$=\underset{x\to 0}{\lim }\frac{1-x^6}{7}+\frac{x^3+x^8}{9}=\frac{1}{7}$$

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