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Question Number 127459 by liberty last updated on 30/Dec/20

 lim_(x→∞) (1+(1/(x+(1/(2+(1/x))))) )^x^2   =?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{{x}}}}\:\right)^{{x}^{\mathrm{2}} } \:=?\: \\ $$

Answered by john_santu last updated on 30/Dec/20

1+(1/(x+(1/(2+(1/x))))) = 1+(1/(x+(x/(2x+1)))) = 1+(1/((2x^2 +2x)/(2x+1)))   ⇒ lim_(x→∞)  (1+(1/((2x^2 +2x)/(2x+1))))^x^2   =         e^(lim_(x→∞) (1+(1/((2x^2 +2x)/(2x+1))) −1 ). x^2 ) =        e^(lim_(x→∞) (((2x+1)/(2x^2 +2x))).x^2 ) = e^∞  = ∞

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2x}+\mathrm{1}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2x}+\mathrm{1}}} \\ $$$$\:\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2x}+\mathrm{1}}}\right)^{\mathrm{x}^{\mathrm{2}} } \:=\: \\ $$$$\:\:\:\:\:\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2x}+\mathrm{1}}}\:−\mathrm{1}\:\right).\:\mathrm{x}^{\mathrm{2}} } = \\ $$$$\:\:\:\:\:\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}\right).\mathrm{x}^{\mathrm{2}} } =\:\mathrm{e}^{\infty} \:=\:\infty\: \\ $$$$ \\ $$

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