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Question Number 78325 by jagoll last updated on 16/Jan/20

lim_(x→∞) (((5x^2 +3x)/(5x^2 −2x)))^(3x−1)

$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \\ $$

Commented by john santu last updated on 16/Jan/20

consider lim_(x→∞) (1+(1/((((5x^2 −2x)/(5x))))))^((5x^2 −2x)/(5x)) =e

$${consider}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{5}{x}}\right)}\right)^{\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{5}{x}}} ={e} \\ $$

Commented by john santu last updated on 16/Jan/20

by short cut e^(lim_(x→∞) (3x−1)(((5x)/(5x^2 −2x)))) = e^3

$${by}\:{short}\:{cut}\: \\ $$$${e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)} =\:{e}^{\mathrm{3}} \\ $$

Commented by mathmax by abdo last updated on 16/Jan/20

let A(x)=(((5x^2 +3x)/(5x^2 −2x)))^(3x−1) ⇒A(x)=e^((3x−1)ln(((5x^2 +3x)/(5x^2 −2x)))) we have ln(((5x^2 +3x)/(5x^2 −2x))) =ln(((5x^2 −2x+5x)/(5x^2 −2x)))=ln(1+((5x)/(5x^2 −2x))) ∼((5x)/(5x^2 −2x)) ∼(1/x)(x→+∞) ⇒A(x)∼e^((3x−1)/x) =e^3 .e^(−(1/x)) ⇒ lim_(x→+∞) A(x)=e^3

$${let}\:{A}\left({x}\right)=\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \:\Rightarrow{A}\left({x}\right)={e}^{\left(\mathrm{3}{x}−\mathrm{1}\right){ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)} \\ $$$${we}\:{have}\:{ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)\:={ln}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)={ln}\left(\mathrm{1}+\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right) \\ $$$$\sim\frac{\mathrm{5}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\:\sim\frac{\mathrm{1}}{{x}}\left({x}\rightarrow+\infty\right)\:\Rightarrow{A}\left({x}\right)\sim{e}^{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}}} \:={e}^{\mathrm{3}} .{e}^{−\frac{\mathrm{1}}{{x}}} \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)={e}^{\mathrm{3}} \\ $$

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