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Question Number 119867 by Study last updated on 27/Oct/20

$$li\underset{x\to \mathrm{\infty }}{m}\frac{{(1+\frac{1}{2n})}^n-\sqrt{e}}{{(1+\frac{2}{n})}^n-e^2}=???$$

Commented by Dwaipayan Shikari last updated on 27/Oct/20

$$\frac{1}{16}{e}^{-\frac{3}{2}}$$

Commented by Ar Brandon last updated on 27/Oct/20

$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{{(1+\frac{1}{2\mathrm{n}})}^{\mathrm{n}}-\sqrt{\mathrm{e}}}{{(1+\frac{2}{\mathrm{n}})}^{\mathrm{n}}-{\mathrm{e}}^2}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{{(1+\frac{1}{2\mathrm{n}})}^{2\mathrm{n}\cdot \frac{1}{2}}-\sqrt{\mathrm{e}}}{{(1+\frac{2}{\mathrm{n}})}^{\frac{\mathrm{n}}{2}\cdot 2}-{\mathrm{e}}^2}$$$$=\frac{\sqrt{\mathrm{e}}-\sqrt{\mathrm{e}}}{{\mathrm{e}}^2-{\mathrm{e}}^2}\{\mathrm{since}\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{(1+\frac{1}{\mathrm{x}})}^{\mathrm{x}}=\mathrm{e}\}$$$$=\frac{\sqrt{\mathrm{e}}-\sqrt{\mathrm{e}}}{(\sqrt{\mathrm{e}}-\sqrt{\mathrm{e}})(\sqrt{\mathrm{e}}+\sqrt{\mathrm{e}})(\mathrm{e}+\mathrm{e})}=\frac{1}{2\sqrt{\mathrm{e}}\times 2\mathrm{e}}=\frac{1}{4}{\mathrm{e}}^{-3/2}$$$$$$$$GreetingstomrRasheed.{It}^{\prime }sbeenquite$$$$alongtime.Hope{you}^{\prime }redoingwellSir!$$

Answered by TANMAY PANACEA last updated on 27/Oct/20

$$y=\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{2n})}^n$$$$lny=\underset{n\to \mathrm{\infty }}{\lim }nln(1+\frac{1}{2n})$$$$=\underset{h\to 0}{\lim }\frac{1}{h}ln(1+\frac{h}{2})$$$$=\underset{h\to 0}{\lim }\frac{ln(1+\frac{h}{2})}{\frac{h}{2}\times 2}=\frac{1}{2}$$$$y=e^{\frac{1}{2}}$$$$similarly$$$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{2}{n})}^n=p$$$$lnp=\underset{n\to \mathrm{\infty }}{\lim }nln(1+\frac{2}{n})$$$$=\underset{h\to 0}{\lim }\frac{ln(1+2h)}{2h}\times 2=2$$$$p=e^2$$$$y=\sqrt{e}\to y^2=eso\mathit{p}={\mathit{y}}^4$$$$\underset{h\to 0}{\lim }\frac{y-\sqrt{e}}{p-e^2}=\underset{h\to 0}{\lim }\frac{\mathit{y}-\sqrt{\mathit{e}}}{{\mathit{y}}^4-{\mathit{e}}^2}$$$$\underset{h\to 0}{\lim }\frac{\sqrt{{\mathit{y}}^2}-\sqrt{\mathit{e}}}{{\left({\mathit{y}}^2\right)}^2-{\left(\mathit{e}\right)}^2}=\underset{\mathit{h}\to 0}{\lim }=\frac{\sqrt{{\mathit{y}}^2}-\sqrt{\mathit{e}}}{({\mathit{y}}^2+\mathit{e})(\sqrt{{\mathit{y}}^2}+\sqrt{\mathit{e}})(\sqrt{{\mathit{y}}^2}-\sqrt{\mathit{e}})}$$$$=\frac{1}{(\mathit{e}+\mathit{e})(\sqrt{\mathit{e}}+\sqrt{\mathit{e}})}=\frac{1}{4}\times \frac{1}{{\mathit{e}}^{\frac{3}{2}}}$$$$$$

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