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Question Number 148439 by liberty last updated on 28/Jul/21
lim_(x→2) (((√(x+2)) ((x+6))^(1/3) −x^2 )/(x−2)) =?
$$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{6}}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{2}}\:=? \\ $$
Answered by dumitrel last updated on 28/Jul/21
lim_(x→2) (((√(x+2))(((x+6))^(1/3) −2))/(x−2))+lim_(x→2) ((2((√(x+2))−2))/(x−2))−lim_(x→2) ((x^2 −4)/(x−2))= lim_(x→2) (((√(x+2))(x−2))/((x−2)((((x+6)^2 ))^(1/3) +2((x+2))^(1/3) +4))+lim_(x→2) ((2(x−2))/((x−2)((√(x+2))+2)))+lim_(x→2) (((x−2)(x+2))/(x−2)))= =(2/(4+4+4))+(2/(2+2))−4=−((10)/3)
$$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\sqrt{{x}+\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{6}}−\mathrm{2}\right)}{{x}−\mathrm{2}}+\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{2}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)}{{x}−\mathrm{2}}−\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}−\mathrm{2}}= \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\sqrt{{x}+\mathrm{2}}\left({x}−\mathrm{2}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{6}\right)^{\mathrm{2}} }+\mathrm{2}\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}+\mathrm{4}\right.}+\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{2}\left({x}−\mathrm{2}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}+\mathrm{2}}+\mathrm{2}\right)}+\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}{\left.{x}−\mathrm{2}\right)}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{4}+\mathrm{4}+\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{2}+\mathrm{2}}−\mathrm{4}=−\frac{\mathrm{10}}{\mathrm{3}} \\ $$
Answered by EDWIN88 last updated on 28/Jul/21
Answered by mathmax by abdo last updated on 28/Jul/21
f(x)=(((√(x+2))(x+6)^(1/3) −x^2 )/(x−2)) changement x−2=t give f(x)=g(t)=(((√(t+4))(t+8)^(1/3) −(t+2)^2 )/t) =(((√(t+4))(t+8)^(1/3) −t^2 −4t−4)/t)=((2(√(1+(t/4)))(^3 (√8)(1+(t/8))^(1/3) −t^2 −4t−4)/t) ⇒g(t)∼((4(1+(t/8))(1+(t/(24)))−t^2 −4t−4)/t) =((4(1+(t/(24))+(t/8)+(t^2 /(8.24)))−t^2 −4t−4)/t) =(((t/6)+(t/2)+(t^2 /(48))−t^2 −4t)/t)=(1/6)+(1/2)−4+((t/(48))−t) ⇒ lim_(t→0) g(t)=(2/3)−4 =−((10)/3) ⇒lim_(x⌣2) f(x)=−((10)/3)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}+\mathrm{2}}\left(\mathrm{x}+\mathrm{6}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{2}}\:\mathrm{changement}\:\mathrm{x}−\mathrm{2}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{\sqrt{\mathrm{t}+\mathrm{4}}\left(\mathrm{t}+\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{t}} \\ $$$$=\frac{\sqrt{\mathrm{t}+\mathrm{4}}\left(\mathrm{t}+\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{4}}{\mathrm{t}}=\frac{\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{4}}}\left(^{\mathrm{3}} \sqrt{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{8}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{4}\right.}{\mathrm{t}} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{8}}\right)\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{24}}\right)−\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{4}}{\mathrm{t}} \\ $$$$=\frac{\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{24}}+\frac{\mathrm{t}}{\mathrm{8}}+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}.\mathrm{24}}\right)−\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{4}}{\mathrm{t}} \\ $$$$=\frac{\frac{\mathrm{t}}{\mathrm{6}}+\frac{\mathrm{t}}{\mathrm{2}}+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{48}}−\mathrm{t}^{\mathrm{2}} −\mathrm{4t}}{\mathrm{t}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{4}+\left(\frac{\mathrm{t}}{\mathrm{48}}−\mathrm{t}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{4}\:=−\frac{\mathrm{10}}{\mathrm{3}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\smile\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{10}}{\mathrm{3}} \\ $$

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