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Question Number 112900 by bemath last updated on 10/Sep/20

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

Commented by kaivan.ahmadi last updated on 10/Sep/20

$$\sim {lim}_{x\to \mathrm{\infty }}{\left(\frac{x^3}{2x^3}\right)}^x={\left(\frac{1}{2}\right)}^{+\mathrm{\infty }}=0$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{{\mathrm{x}}^3-{\mathrm{x}}^2+1}{{2\mathrm{x}}^3+{\mathrm{x}}^2-2}\right)}^{\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\left\{\frac{1-\frac{1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}}{2+\frac{1}{\mathrm{x}}-\frac{2}{{\mathrm{x}}^3}}\right\}}^{\mathrm{x}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{xln}\left(\frac{1-\frac{1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}}{2+\frac{1}{\mathrm{x}}-\frac{2}{{\mathrm{x}}^3}}\right)}\sim {\mathrm{e}}^{-\mathrm{xln}2}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=0\mathrm{and}$$$${\lim }_{\mathrm{x}\to -\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$

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