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Question Number 3048 by Syaka last updated on 03/Dec/15

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

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

Answered by 123456 last updated on 04/Dec/15

L=lim_(x→−∞) (1+(1/x))^x   u=(1/x)  x→−∞≡u→0^−   L=lim_(u→0^− ) (1+u)^(1/u)   remember that  e=lim_(x→0) (1+x)^(1/x)   so  L=e

$$\mathrm{L}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${u}=\frac{\mathrm{1}}{{x}} \\ $$$${x}\rightarrow−\infty\equiv{u}\rightarrow\mathrm{0}^{−} \\ $$$$\mathrm{L}=\underset{{u}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{{u}}} \\ $$$$\mathrm{remember}\:\mathrm{that} \\ $$$${e}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{so} \\ $$$$\mathrm{L}={e} \\ $$

Commented by prakash jain last updated on 04/Dec/15

y=(1+x)^(1/x)   ln y=(1/x)ln (1+x)  lim_(x→0) ln y=lim_(x→0) ((ln (1+x))/x)=lim_(x→0) (((d/dx)ln (1+x))/((d/dx)x))=lim_(x→0) (1/(1+x))=1  lim_(x→0) ln y=1  lim_(x→0) y=e^1 =e

$${y}=\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\ $$$$\mathrm{ln}\:{y}=\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{\frac{{d}}{{dx}}{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+{x}}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:{y}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{y}={e}^{\mathrm{1}} ={e} \\ $$

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