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lim-x-1-1-x-x-2-e-x-




Question Number 118950 by benjo_mathlover last updated on 21/Oct/20
  lim_(x→∞)  (1+(1/x))^x^2  .e^(−x)  = ?.
$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } .{e}^{−{x}} \:=\:?. \\ $$
Answered by john santu last updated on 21/Oct/20
 L = lim_(x→∞)  (((1+(1/x))^x^2  )/e^x )    ln L= lim_(x→∞)  ln ((((1+(1/x))^x^2  )/e^x ))   ln L = lim_(x→∞)  x^2 (ln (1+(1/x))−x)   [ by using Maclaurin series ]   ln L = lim_(x→∞)  x^2 ((1/x)−(1/(2x^2 ))+(1/(3x^3 ))−O(x^3 ))−x)   ln L = lim_(x→∞)   [x−(1/(2x))+(1/(3x^2 ))−O(x^2 )−x ]  ln L = lim_(x→∞)  (−(1/2)+(1/(3x^2 ))−O(x^2 ))=−(1/2)  L = e^(−(1/2))  = (1/( (√e))) .
$$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } }{{e}^{{x}} }\: \\ $$$$\:\mathrm{ln}\:\mathcal{L}=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\:\left(\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } }{{e}^{{x}} }\right) \\ $$$$\:\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left(\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\right) \\ $$$$\:\left[\:{by}\:{using}\:{Maclaurin}\:{series}\:\right]\: \\ $$$$\left.\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }−{O}\left({x}^{\mathrm{3}} \right)\right)−{x}\right) \\ $$$$\:\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\left[{x}−\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }−{O}\left({x}^{\mathrm{2}} \right)−{x}\:\right] \\ $$$$\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }−{O}\left({x}^{\mathrm{2}} \right)\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathcal{L}\:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt{{e}}}\:. \\ $$$$ \\ $$

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