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Question Number 186617 by mustafazaheen last updated on 07/Feb/23

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2-1}{\sqrt[5]{(x^2-1)}}=?withsolution$$

Commented by aba last updated on 07/Feb/23

$$$$$$=\underset{x\to \mathrm{\infty }}{\lim }\sqrt[5]{\frac{{({\mathrm{x}}^2-1)}^5}{({\mathrm{x}}^2-1)}}=\underset{x\to \mathrm{\infty }}{\lim }\sqrt[5]{{({\mathrm{x}}^2-1)}^4}=\mathrm{\infty }$$

Commented by aba last updated on 07/Feb/23

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1-\frac{1}{{\mathrm{x}}^2}}{\sqrt[5]{\frac{1}{{\mathrm{x}}^8}-\frac{1}{{\mathrm{x}}^{10}}}}=\frac{1}{0}=\mathrm{\infty }$$

Answered by cortano1 last updated on 07/Feb/23

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[5]{{(x^2-1)}^4}=\mathrm{\infty }$$

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