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lim-x-x-x-2-x-1-ln-e-x-x-x-




Question Number 79565 by jagoll last updated on 26/Jan/20
lim_(x→∞) (x−(√(x^2 −x+1)) )×(((ln(e^x +x))/x))
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\:\right)×\left(\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{x}\right)}{\mathrm{x}}\right) \\ $$
Commented by john santu last updated on 26/Jan/20
(1)lim_(x→∞)  (√(x^2  )) −(√(x^2 −x+1)) =((0−(−1))/2)=(1/2)  (2) lim_(x→∞)  ((ln(e^x +x))/x)=lim_(x→∞)  ((e^x +1)/(e^x +x))=  lim_(x→∞)  (e^x /e^x ) = 1  ∴ (1/2)×1 = (1/2)
$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}^{\mathrm{2}} \:}\:−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\:=\frac{\mathrm{0}−\left(−\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{x}\right)}{\mathrm{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}{\mathrm{e}^{\mathrm{x}} +\mathrm{x}}= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} }\:=\:\mathrm{1} \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 26/Jan/20
let f(x)=(x−(√(x^2 −x+1)))(((ln(e^x  +x))/x)) ⇒for x>0  f(x)=(x−(√(x^2 (1−(1/x) +(1/x^2 ))))(((ln(e^x (1+xe^(−x) ))/x))  =(x−x(√(1+(1/x^2 )−(1/x))))(1+((ln(1+xe^(−x) ))/x))  ∼(x−x(1+(1/2)((1/x^2 )−(1/x)))(1+((xe^(−x) )/x))  =(−(1/(2x)) +(1/2))(1 +e^(−x) ) ⇒lim_(x→+∞) f(x)=(1/2)
$${let}\:{f}\left({x}\right)=\left({x}−\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right)\left(\frac{{ln}\left({e}^{{x}} \:+{x}\right)}{{x}}\right)\:\Rightarrow{for}\:{x}>\mathrm{0} \\ $$$${f}\left({x}\right)=\left({x}−\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right.}\right)\left(\frac{{ln}\left({e}^{{x}} \left(\mathrm{1}+{xe}^{−{x}} \right)\right.}{{x}}\right) \\ $$$$=\left({x}−{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}}\right)\left(\mathrm{1}+\frac{{ln}\left(\mathrm{1}+{xe}^{−{x}} \right)}{{x}}\right) \\ $$$$\sim\left({x}−{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}\right)\right)\left(\mathrm{1}+\frac{{xe}^{−{x}} }{{x}}\right)\right. \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}\:+{e}^{−{x}} \right)\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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