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Question-211895




Question Number 211895 by Spillover last updated on 23/Sep/24
Answered by mehdee7396 last updated on 23/Sep/24
lim_(x→∞)  (1+(a/x)−(4/x^2 )−1)2x  =lim_(x→∞)  (((ax−4)/x^2 ))2x=2a  ⇒lim_(x→∞)  (1+(a/x)−(4/x^2 )−1)^(2x) =e^(2a)   ⇒2a=3⇒a=(3/2) ✓
$${lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\mathrm{2}{x} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\left(\frac{{ax}−\mathrm{4}}{{x}^{\mathrm{2}} }\right)\mathrm{2}{x}=\mathrm{2}{a} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)^{\mathrm{2}{x}} ={e}^{\mathrm{2}{a}} \\ $$$$\Rightarrow\mathrm{2}{a}=\mathrm{3}\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$
Answered by BHOOPENDRA last updated on 23/Sep/24
we know if we have function  lim_(x→a)  f(x)^(g(x)) =1^∞   then we use the property   lim_(e^x →a)  g(x) {f(x)−1}  So lim_(e^x →∞)  2x (1+(a/x)−(4/x^2 ) −1)=e^3   lim_(e^x →∞)  2x ((a/x)−(4/x^2 ))=e^3     e^(2(a−(4/∞))) =e^3   e^(2a) =e^3   2a=3  a=(3/2)
$${we}\:{know}\:{if}\:{we}\:{have}\:{function} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)^{{g}\left({x}\right)} =\mathrm{1}^{\infty} \\ $$$${then}\:{we}\:{use}\:{the}\:{property}\: \\ $$$$\underset{{e}^{{x}} \rightarrow{a}} {\mathrm{lim}}\:{g}\left({x}\right)\:\left\{{f}\left({x}\right)−\mathrm{1}\right\} \\ $$$${So}\:\underset{{e}^{{x}} \rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\:−\mathrm{1}\right)={e}^{\mathrm{3}} \\ $$$$\underset{{e}^{{x}} \rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:\left(\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)={e}^{\mathrm{3}} \\ $$$$ \\ $$$${e}^{\mathrm{2}\left({a}−\frac{\mathrm{4}}{\infty}\right)} ={e}^{\mathrm{3}} \\ $$$${e}^{\mathrm{2}{a}} ={e}^{\mathrm{3}} \\ $$$$\mathrm{2}{a}=\mathrm{3} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 24/Sep/24
thanks
$${thanks} \\ $$

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