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Question Number 112974 by bemath last updated on 10/Sep/20

$$\mathrm{solve}\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{{2\mathrm{x}}^3-{\mathrm{x}}^2+2}{{2\mathrm{x}}^3-{4\mathrm{x}}^2+1}\right)}^{\mathrm{x}}$$

Answered by john santu last updated on 10/Sep/20

Answered by mathmax by abdo last updated on 11/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{{2\mathrm{x}}^3-{\mathrm{x}}^2+2}{{2\mathrm{x}}^3-{4\mathrm{x}}^2+1}\right)}^{\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\left(\frac{1-\frac{1}{2\mathrm{x}}+\frac{1}{{\mathrm{x}}^3}}{1-\frac{2}{\mathrm{x}}+\frac{1}{{2\mathrm{x}}^3}}\right)}^{\mathrm{x}}$$$$={\mathrm{e}}^{\mathrm{xln}\left(\frac{1-\frac{1}{2\mathrm{x}}+\frac{1}{{\mathrm{x}}^3}}{1-\frac{2}{\mathrm{x}}+\frac{1}{{2\mathrm{x}}^3}}\right)}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{1}{\mathrm{t}}\right)={\mathrm{e}}^{\frac{1}{\mathrm{t}}\ln \left(\frac{1-\frac{\mathrm{t}}{2}+{\mathrm{t}}^3}{1-2\mathrm{t}+\frac{1}{2}{\mathrm{t}}^3}\right)}(\mathrm{t}\to 0)$$$$={\mathrm{e}}^{\frac{1}{\mathrm{t}}\ln \left(\frac{1-2\mathrm{t}+\frac{{\mathrm{t}}^3}{2}-\frac{\mathrm{t}}{2}+{\mathrm{t}}^3+2\mathrm{t}-\frac{{\mathrm{t}}^3}{2}}{1-2\mathrm{t}+\frac{{\mathrm{t}}^3}{2}}\right)}={\mathrm{e}}^{\frac{1}{\mathrm{t}}\ln (1+\frac{\frac{{\mathrm{t}}^3}{2}+\frac{3}{2}\mathrm{t}}{1-2\mathrm{t}+\frac{{\mathrm{t}}^3}{2}})}$$$$\sim {\mathrm{e}}^{\frac{1}{\mathrm{t}}\left(\frac{\frac{{\mathrm{t}}^3}{2}+\frac{3}{2}\mathrm{t}}{1-2\mathrm{t}+\frac{{\mathrm{t}}^3}{2}}\right)}={\mathrm{e}}^{\frac{{\mathrm{t}}^2+3}{2-4\mathrm{t}+{\mathrm{t}}^3}}\to {\mathrm{e}}^{\frac{3}{2}}(\mathrm{t}\to 0)\Rightarrow {\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}\sqrt{\mathrm{e}}$$

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