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Question Number 118950 by benjo_mathlover last updated on 21/Oct/20

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

Answered by john santu last updated on 21/Oct/20

$$\mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }\frac{{(1+\frac{1}{x})}^{x^2}}{e^x}$$$$\ln \mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }\ln \left(\frac{{(1+\frac{1}{x})}^{x^2}}{e^x}\right)$$$$\ln \mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }{x}^2(\ln (1+\frac{1}{x})-x)$$$$\left[byusingMaclaurinseries\right]$$$$\ln \mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }{x}^2(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-O\left(x^3\right))-x)$$$$\ln \mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }[x-\frac{1}{2x}+\frac{1}{3x^2}-O\left(x^2\right)-x]$$$$\ln \mathcal{L}=\underset{x\to \mathrm{\infty }}{\lim }(-\frac{1}{2}+\frac{1}{3x^2}-O\left(x^2\right))=-\frac{1}{2}$$$$\mathcal{L}=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}.$$$$$$

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