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Question Number 92503 by john santu last updated on 07/May/20

lim_(x→∞) (e^(10x) −8x)^(1/(2x)) =

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:=\: \\ $$

Commented by mathmax by abdo last updated on 07/May/20

let f(x)=(e^(10x) −8x)^(1/(2x)) ⇒f(x) =(e^(10x) )^(1/(2x)) (1−8e^(−10x) x)^(1/(2x)) =e^5 (1−8x e^(−10x) )^(1/(2x)) ∼ e^5 (1−8xe^(−10x) ×(1/(2x))) =e^5 (1−4e^(−10x) ) →e^5 (x→+∞) and lim_(x→−∞) f(x) =−∞

$${let}\:{f}\left({x}\right)=\left({e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\Rightarrow{f}\left({x}\right)\:=\left({e}^{\mathrm{10}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \left(\mathrm{1}−\mathrm{8}{e}^{−\mathrm{10}{x}} \:{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$={e}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{8}{x}\:{e}^{−\mathrm{10}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\sim\:{e}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{8}{xe}^{−\mathrm{10}{x}} ×\frac{\mathrm{1}}{\mathrm{2}{x}}\right) \\ $$$$={e}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{4}{e}^{−\mathrm{10}{x}} \right)\:\rightarrow{e}^{\mathrm{5}} \:\:\:\left({x}\rightarrow+\infty\right) \\ $$$${and}\:{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)\:=−\infty \\ $$

Commented by john santu last updated on 07/May/20

f(x)=(e^(10x) −8x)^(1/(2x)) lim_(x→∞) ln f(x)=lim_(x→∞) ln(e^(10x) −8x)^(1/(2x)) lim_(x→∞) ln f(x)= lim_(x→∞) ((ln(e^(10x) −8x))/(2x)) lim_(x→∞) ln f(x) =^H lim_(x→∞) ((2(10e^(10x) −8))/(e^(10x) −8x)) lim_(x→∞) ln f(x)= 2lim_(x→∞) (((10−8e^(−10x) )/(1−8xe^(−10x) ))) lim_(x→∞) ln f(x) = 20 ∴ lim_(x→∞) f(x) = e^(20)

$${f}\left({x}\right)=\left({e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{f}\left(\mathrm{x}\right)=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)}{\mathrm{2}{x}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{f}\left(\mathrm{x}\right)\:\overset{\mathrm{H}} {=}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{10e}^{\mathrm{10}{x}} −\mathrm{8}\right)}{{e}^{\mathrm{10}{x}} −\mathrm{8}{x}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}ln}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{2}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{10}−\mathrm{8e}^{−\mathrm{10}{x}} }{\mathrm{1}−\mathrm{8}{xe}^{−\mathrm{10}{x}} }\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}ln}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{20} \\ $$$$\therefore\:\underset{{x}\rightarrow\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)\:=\:\mathrm{e}^{\mathrm{20}} \: \\ $$

Commented by jagoll last updated on 08/May/20

wrong sir.

$$\mathrm{wrong}\:\mathrm{sir}.\: \\ $$

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