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Question Number 128149 by MathSh last updated on 04/Jan/21

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

Answered by liberty last updated on 04/Jan/21

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

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