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Question Number 130946 by help last updated on 30/Jan/21

Answered by mathmax by abdo last updated on 30/Jan/21

$${\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{{\mathrm{n}}^4-{2\mathrm{n}}^3}-{\mathrm{n}}^2}{\mathrm{n}+2}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{{\mathrm{n}}^4(1-\frac{2}{\mathrm{n}})}-{\mathrm{n}}^2}{\mathrm{n}+2}$$$$=\frac{{\mathrm{n}}^2}{\mathrm{n}+2}(\sqrt{1-\frac{2}{\mathrm{n}}}-1)\Rightarrow {\mathrm{u}}_{\mathrm{n}}\sim \frac{{\mathrm{n}}^2}{\mathrm{n}+2}(1-\frac{1}{\mathrm{n}}-1)=-\frac{\mathrm{n}}{\mathrm{n}+2}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}^3\sim -\frac{{\mathrm{n}}^3}{{(\mathrm{n}+2)}^3}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=-1$$

Answered by malwan last updated on 31/Jan/21

$$\underset{x\to \mathrm{\infty }}{lim}{\left[\frac{n^4-2n^3-n^4}{(n+2)(\sqrt{n^4-2n^3}+n^2)}\right]}^3$$$$=\underset{x\to \mathrm{\infty }}{lim}{\left[\frac{-2n^3}{(n+2)n^2(\sqrt{1-\frac{2}{n}}+1)}\right]}^3$$$$\approx \underset{x\to \mathrm{\infty }}{lim}{\left[\frac{-2n^3}{(n^3+2n^2)\times 2}\right]}^3={\left(\frac{-2}{2}\right)}^3=-1$$

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