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Question Number 73518 by aliesam last updated on 13/Nov/19

Commented by mathmax by abdo last updated on 13/Nov/19

$$letA\left(x\right)={\left(\frac{x+1}{x-1}\right)}^{2x-1}\Rightarrow A\left(x\right)=e^{(2x-1)ln\left(\frac{x+1}{x-1}\right)}$$$$=e^{(2x-1)ln\left(\frac{x-1+2}{x-1}\right)}=e^{(2x-1)ln(1+\frac{2}{x-1})}wehaveln(1+\frac{2}{x-1})\sim \frac{2}{x-1}(x\to +\mathrm{\infty })$$$$\Rightarrow (2x-1)ln(1+\frac{2}{x-1})\sim \frac{2(2x-1)}{x-1}\to 4(x\to +\mathrm{\infty })\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}A\left(x\right)=e^4$$

Answered by ajfour last updated on 13/Nov/19

$$L={\left\{\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{2}{x-1})}^{\left(\frac{x-1}{2}\right)}\right\}}^{\underset{x\to \mathrm{\infty }}{\lim }\frac{2(2x-1)}{x-1}}$$$$L=e^4.$$

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