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Question Number 59393 by Mikael_Marshall last updated on 09/May/19

$$\text{Missing left or extra right}$$$$pls.$$

Commented by maxmathsup by imad last updated on 10/May/19

$$letA\left(x\right)={\left(\frac{x^2-4}{x^2+2}\right)}^{\frac{x^2-1}{x+1}}\Rightarrow ln\left(A\left(x\right)\right)=\left(\frac{x^2-1}{x^2+2}\right)ln\left(\frac{x^2-4}{x^2+2}\right)but$$$$ln\left(\frac{x^2-4}{x^2+2}\right)=ln\left(\frac{x^2+2-6}{x^2+2}\right)=ln(1-\frac{6}{x^2+2})\sim \frac{-6}{x^2+2}(x\to +\mathrm{\infty })(becauseln(1+u)\sim u(u\to 0)$$$$\Rightarrow ln\left(A\left(x\right)\right)\sim \frac{-6(x^2-1)}{{(x^2+2)}^2}\to 0(x\to +\mathrm{\infty })\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=1.$$

Commented by Mikael_Marshall last updated on 10/May/19

$$thankyouSir$$

Commented by mr W last updated on 10/May/19

$$pleasechecksir:$$$$letA\left(x\right)={\left(\frac{x^2-4}{x^2+2}\right)}^{\frac{x^2-1}{x+1}}\Rightarrow ln\left(A\left(x\right)\right)=\left(\frac{x^2-1}{x+1}\right)ln\left(\frac{x^2-4}{x^2+2}\right)$$$$$$

Commented by kaivan.ahmadi last updated on 10/May/19

$$iflimf{\left(x\right)}^{g\left(x\right)}=1^{\mathrm{\infty }}wecanuse$$$$lim{e}^{g\left(x\right)(f\left(x\right)-1)}$$$$so$$$${lim}_{x\to +\mathrm{\infty }}{e}^{\left(\frac{x^2-1}{x+1}\right)(\frac{x^2-4}{x^2+2}-1)}={lim}_{x\to +\mathrm{\infty }}{e}^{\left(\frac{x^2-1}{x+1}\right)\left(\frac{-6}{x^2+2}\right)}=$$$$e^0=1$$

Commented by Mikael_Marshall last updated on 10/May/19

$$thanksSir$$

Commented by maxmathsup by imad last updated on 10/May/19

$$yessirihavecommitedaerrorletrectifyln\left(A\left(x\right)\right)=\frac{x^2-1}{x+1}ln\left(\frac{x^2-4}{x^2+2}\right)\Rightarrow$$$$ln\left(A\left(x\right)\right)\sim \frac{x^2-1}{x+1}.\frac{-6}{x^2+2}=\frac{-6x^2+6}{(x+1)(x^2+2)}\to 0(x\to +\mathrm{\infty })\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=1.$$$$$$

Answered by malwaan last updated on 10/May/19

$$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x^2+2-6}{x^2+2}\right)}^{\frac{(x+1)(x-1)}{x+1}}$$$$=\underset{x\to \mathrm{\infty }}{lim}{(1-\frac{6}{x^2+2})}^{x-1}$$$$=\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{-6}{x^2+2})}^{\left(\frac{x^2+2}{-6}\right)(x-1)\left(\frac{-6}{x^2+2_{}}\right)}$$$$=\underset{x\to \mathrm{\infty }}{lim}{\left[{(1+\frac{-6}{x^2+2})}^{\left(\frac{x^2+2}{-6}\right)}\right]}^{\frac{-6(x-1)}{x^2+2}}$$$$={\left[e^{-6}\right]}^0=1$$

Commented by Mikael_Marshall last updated on 10/May/19

$$thanksSir$$

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