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Lim-x-4x-2-8x-3-2x-4-




Question Number 82541 by zainal tanjung last updated on 22/Feb/20
 Lim_(x→∞)  (√(4x^2 −8x+3)) −2x−4
$$\:\underset{\mathrm{x}\rightarrow\infty} {\mathrm{Lim}}\:\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{3}}\:−\mathrm{2x}−\mathrm{4} \\ $$
Commented by john santu last updated on 22/Feb/20
lim_(x→∞)  (√(4x^2 −8x+3)) −(2x+4) =  lim_(x→∞)  (√(4x^2 −8x+3)) −(√(4x^2 +16x+16)) =  ((−8−16)/(2.2)) = −6
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}\:−\left(\mathrm{2}{x}+\mathrm{4}\right)\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{16}}\:= \\ $$$$\frac{−\mathrm{8}−\mathrm{16}}{\mathrm{2}.\mathrm{2}}\:=\:−\mathrm{6}\: \\ $$
Commented by mathmax by abdo last updated on 22/Feb/20
lim_(x→−∞) (√(4x^2 −8x+3))−2x−4 =+∞  at +∞  let f(x)=(√(4x^2 −8x+3))−2x−4  we have f(x)=2x(√(1−(2/x)+(3/(4x^2 ))))−2x−4  ∼2x(1+(1/2)(−(2/x)+(3/(4x^2 ))))−2x−4 =2x−2+(3/(4x))−2x−4 ⇒  f(x)∼(3/(4x)) −6 ⇒lim_(x→+∞) f(x)=−6
$${lim}_{{x}\rightarrow−\infty} \sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}−\mathrm{2}{x}−\mathrm{4}\:=+\infty \\ $$$${at}\:+\infty\:\:{let}\:{f}\left({x}\right)=\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}}−\mathrm{2}{x}−\mathrm{4} \\ $$$${we}\:{have}\:{f}\left({x}\right)=\mathrm{2}{x}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }}−\mathrm{2}{x}−\mathrm{4} \\ $$$$\sim\mathrm{2}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }\right)\right)−\mathrm{2}{x}−\mathrm{4}\:=\mathrm{2}{x}−\mathrm{2}+\frac{\mathrm{3}}{\mathrm{4}{x}}−\mathrm{2}{x}−\mathrm{4}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{3}}{\mathrm{4}{x}}\:−\mathrm{6}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=−\mathrm{6} \\ $$

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