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Question Number 182978 by TUN last updated on 18/Dec/22

Answered by dumitrel last updated on 18/Dec/22

Commented by dumitrel last updated on 18/Dec/22

Answered by cortano1 last updated on 18/Dec/22

$$L=\underset{n\to \mathrm{\infty }}{\lim }(\sqrt{n^2+n}.\sqrt[3]{n^3-n^2}-n^2)$$$$L=\underset{n\to \mathrm{\infty }}{\lim }{n}^2\sqrt{1+\frac{1}{n}}.\sqrt[3]{1-\frac{1}{n}}-n^2$$$$[let\frac{1}{n}=x]$$$$L=\underset{x\to 0}{\lim }\frac{\sqrt{1+x}.\sqrt[3]{1-x}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt[6]{{(1+x)}^3}.\sqrt[6]{{(1-x)}^2}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt[6]{(1+x){(1-x^2)}^2}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt[6]{(1+x)(1-2x^2+x^4)}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt[6]{1+x^5+x^4-2x^3-2x^2+x}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{1+\left(\frac{x^5+x^4-2x^3-2x^2+x}{6}\right)-1}{x^2}=\mathrm{\infty }$$

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