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lim-x-x-a-x-a-x-




Question Number 111442 by bobhans last updated on 03/Sep/20
  lim_(x→∞)  (((x+a)/(x−a)))^x ?
$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}−\mathrm{a}}\right)^{\mathrm{x}} ? \\ $$
Answered by bemath last updated on 03/Sep/20
Answered by ajfour last updated on 03/Sep/20
ln L=lim_(x→∞) xln (((1+(a/x))/(1−(a/x))))         = lim_(x→∞) {xln (1+(a/x))−xln (1−(a/x))}         = alim_(x→∞) {((ln (1+(a/x)))/(((a/x))))+((ln (1−(a/x)))/((−(a/x))))}  ln L = 2a  L = e^(2a)  .
$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\mathrm{ln}\:\left(\frac{\mathrm{1}+\frac{{a}}{{x}}}{\mathrm{1}−\frac{{a}}{{x}}}\right) \\ $$$$\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{{x}\mathrm{ln}\:\left(\mathrm{1}+\frac{{a}}{{x}}\right)−{x}\mathrm{ln}\:\left(\mathrm{1}−\frac{{a}}{{x}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:=\:{a}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{{a}}{{x}}\right)}{\left(\frac{{a}}{{x}}\right)}+\frac{\mathrm{ln}\:\left(\mathrm{1}−\frac{{a}}{{x}}\right)}{\left(−\frac{{a}}{{x}}\right)}\right\} \\ $$$$\mathrm{ln}\:{L}\:=\:\mathrm{2}{a} \\ $$$${L}\:=\:{e}^{\mathrm{2}{a}} \:. \\ $$
Answered by Dwaipayan Shikari last updated on 03/Sep/20
lim_(x→∞) (1+((2a)/(x−a)))^((x/(2a)).((x−a)/(x−a)).2a)   =e^((2ax)/(x−a)) =e^(2a)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{2}{a}}{{x}−{a}}\right)^{\frac{{x}}{\mathrm{2}{a}}.\frac{{x}−{a}}{{x}−{a}}.\mathrm{2}{a}} \\ $$$$={e}^{\frac{\mathrm{2}{ax}}{{x}−{a}}} ={e}^{\mathrm{2}{a}} \\ $$

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