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Question Number 160241 by mathlove last updated on 26/Nov/21

Commented by blackmamba last updated on 26/Nov/21

$$\underset{x\to 1}{\lim }\frac{\sqrt[9]{7x-6}-1}{{nx}^3-n}$$$$=\frac{1}{n}\underset{x\to 1}{\lim }\frac{\sqrt[9]{7x-6}-1}{(x^2+x+1)(x-1)}$$$$=\frac{1}{3n}\underset{x\to 1}{\lim }\frac{\sqrt[9]{7x-6}-1}{x-1}$$$$let7x-6=u^9$$$$L=\frac{1}{3n}\underset{u\to 1}{\lim }\frac{u-1}{\left(\frac{u^9+6}{7}\right)-1}$$$$L=\frac{7}{3n}\underset{u\to 1}{\lim }\frac{u-1}{u^9-1}$$$$L=\frac{7}{3n}\times \frac{1}{9}=\frac{7}{27n}$$

Answered by qaz last updated on 26/Nov/21

$$\underset{\mathrm{x}\to 1}{\lim }\frac{\sqrt[9]{7\mathrm{x}-6}-1}{{\mathrm{nx}}^3-\mathrm{n}}=\frac{1}{\mathrm{n}}\underset{\mathrm{x}\to 1,\xi \to 1}{\lim }\frac{{\xi }^{-8/9}(7\mathrm{x}-7)}{9({\mathrm{x}}^3-1)}=\frac{7}{9\mathrm{n}}\underset{\mathrm{x}\to 1}{\lim }\frac{1}{{\mathrm{x}}^2+\mathrm{x}+1}=\frac{7}{27\mathrm{n}}$$

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