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lim-x-e-x-cos-x-ln-x-2-2x-3-x-




Question Number 81558 by john santu last updated on 14/Feb/20
lim_(x→∞)  ((e^x  +cos x+ln(x^2 −2x+3))/x) =
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{cos}\:\mathrm{x}+\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{3}\right)}{\mathrm{x}}\:=\: \\ $$
Commented by malwaan last updated on 14/Feb/20
∞
$$\infty \\ $$
Commented by mathmax by abdo last updated on 14/Feb/20
f(x)=((e^x +cosx +ln(x^2 −2x +3))/x)  ⇒for x→+∞  f(x)=((e^x  +cosx+ 2lnx+ln(1−(2/x) +(3/x^2 )))/x)  ∼((e^x  +cosx+2lnx +(−(2/x)+(3/x^2 )))/x) =(e^x /x) +((cosx)/x) +((2lnx)/x)−(2/x^2 ) +(3/x^3 )  now its clear that lim_(x→+∞)  f(x)=+∞  for −∞   f(x) ∼(e^x /x) +((cosx)/x) +((2ln(−x))/x)−(2/x^2 ) +(3/x^3 ) ⇒  lim_(x→−∞) f(x)=0
$${f}\left({x}\right)=\frac{{e}^{{x}} +{cosx}\:+{ln}\left({x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{3}\right)}{{x}}\:\:\Rightarrow{for}\:{x}\rightarrow+\infty \\ $$$${f}\left({x}\right)=\frac{{e}^{{x}} \:+{cosx}+\:\mathrm{2}{lnx}+{ln}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}{{x}} \\ $$$$\sim\frac{{e}^{{x}} \:+{cosx}+\mathrm{2}{lnx}\:+\left(−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}{{x}}\:=\frac{{e}^{{x}} }{{x}}\:+\frac{{cosx}}{{x}}\:+\frac{\mathrm{2}{lnx}}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}^{\mathrm{3}} } \\ $$$${now}\:{its}\:{clear}\:{that}\:{lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)=+\infty \\ $$$${for}\:−\infty\:\:\:{f}\left({x}\right)\:\sim\frac{{e}^{{x}} }{{x}}\:+\frac{{cosx}}{{x}}\:+\frac{\mathrm{2}{ln}\left(−{x}\right)}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}^{\mathrm{3}} }\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)=\mathrm{0} \\ $$

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