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Question Number 84600 by john santu last updated on 14/Mar/20
lim_(x→∞ ) x(√((x−1)/(9x+2))) − (x/3)
$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:{x}\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\:\frac{{x}}{\mathrm{3}} \\ $$
Commented by jagoll last updated on 14/Mar/20
nice mister
$$\mathrm{nice}\:\mathrm{mister} \\ $$
Commented by john santu last updated on 14/Mar/20
if lim_(x→0) (√(1−((11)/(9(x+(2/9)))))) ≈ 1−((11)/(18(x+(2/9)))) but lim_(x→∞) (√(1−((11)/(9(x+(2/9)))))) ≇ 1−((11)/(18(x+(2/9)))) your answer not correct sir
$$\mathrm{if}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\mathrm{1}−\frac{\mathrm{11}}{\mathrm{9}\left(\mathrm{x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}}\:\approx\:\mathrm{1}−\frac{\mathrm{11}}{\mathrm{18}\left(\mathrm{x}+\frac{\mathrm{2}}{\mathrm{9}}\right)} \\ $$$$\mathrm{but}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{1}−\frac{\mathrm{11}}{\mathrm{9}\left(\mathrm{x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}}\:\ncong\:\mathrm{1}−\frac{\mathrm{11}}{\mathrm{18}\left(\mathrm{x}+\frac{\mathrm{2}}{\mathrm{9}}\right)} \\ $$$$\mathrm{your}\:\mathrm{answer}\:\mathrm{not}\:\mathrm{correct}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 14/Mar/20
let l(x)=x(√((x−1)/(9x+2)))−(x/3) ⇒l(x)=(x/3)(√((x−1)/(x+(2/9))))−(x/3) =(x/3)(√((x+(2/9)−1−(2/9))/(x+(2/9))))−(x/3) =(x/3)(√(1−((11)/(9(x+(2/9))))))−(x/3) ⇒l(x)∼(x/3)(1−((11)/(18(x+(2/9)))))−(x/x) =−((11x)/(54(x+(2/9)))) ⇒ lim_(x→+∞) f(x) =−((11)/(54))
$${let}\:{l}\left({x}\right)={x}\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}−\frac{{x}}{\mathrm{3}}\:\Rightarrow{l}\left({x}\right)=\frac{{x}}{\mathrm{3}}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\frac{\mathrm{2}}{\mathrm{9}}}}−\frac{{x}}{\mathrm{3}} \\ $$$$=\frac{{x}}{\mathrm{3}}\sqrt{\frac{{x}+\frac{\mathrm{2}}{\mathrm{9}}−\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}}{{x}+\frac{\mathrm{2}}{\mathrm{9}}}}−\frac{{x}}{\mathrm{3}}\:=\frac{{x}}{\mathrm{3}}\sqrt{\mathrm{1}−\frac{\mathrm{11}}{\mathrm{9}\left({x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}}−\frac{{x}}{\mathrm{3}} \\ $$$$\Rightarrow{l}\left({x}\right)\sim\frac{{x}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{11}}{\mathrm{18}\left({x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}\right)−\frac{{x}}{{x}}\:=−\frac{\mathrm{11}{x}}{\mathrm{54}\left({x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:\:{f}\left({x}\right)\:=−\frac{\mathrm{11}}{\mathrm{54}} \\ $$
Commented by abdomathmax last updated on 14/Mar/20
no sir put x+(2/9)=t (so t→+∞) ⇒ (√(1−((11)/(9(x+(2/9))))))=(√(1−((11)/(9t))))∼1−((11)/(18t)) there is no error sir...
$${no}\:{sir}\:{put}\:\:{x}+\frac{\mathrm{2}}{\mathrm{9}}={t}\:\:\left({so}\:{t}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{1}−\frac{\mathrm{11}}{\mathrm{9}\left({x}+\frac{\mathrm{2}}{\mathrm{9}}\right)}}=\sqrt{\mathrm{1}−\frac{\mathrm{11}}{\mathrm{9}{t}}}\sim\mathrm{1}−\frac{\mathrm{11}}{\mathrm{18}{t}} \\ $$$${there}\:{is}\:{no}\:{error}\:{sir}… \\ $$
Commented by abdomathmax last updated on 14/Mar/20
let try another way let f(x)=x((√((x−1)/(9x+2)))−(1/3)) we do the changement (√((x−1)/(9x+2)))=t ⇒ ((x−1)/(9x+2))=t^2 ⇒x−1=9t^2 x +2t^2 ⇒(1−9t^2 )x=1+2t^2 ⇒ x=((1+2t^2 )/(1−9t^2 )) we have x→+∞ ⇒t→(1/3) and f(x)=(((1+2t^2 )/(1−9t^2 )))(t−(1/3))=B(t) ⇒ B(t)=(((3t−1)(2t^2 +1))/(3(1−3t)(1+3t))) ⇒B(t)=−(1/3)×((2t^2 +1)/(3t+1)) ⇒ lim_(t→(1/3)) B(t) =−(1/3)×(((2/9)+1)/2) =−(1/6)(((11)/9)) =−((11)/(54))
$${let}\:{try}\:{another}\:{way}\:\:{let}\:{f}\left({x}\right)={x}\left(\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${we}\:{do}\:{the}\:{changement}\:\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}={t}\:\Rightarrow \\ $$$$\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}={t}^{\mathrm{2}} \:\Rightarrow{x}−\mathrm{1}=\mathrm{9}{t}^{\mathrm{2}} {x}\:+\mathrm{2}{t}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} \right){x}=\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\:\:\:{we}\:{have}\:{x}\rightarrow+\infty\:\Rightarrow{t}\rightarrow\frac{\mathrm{1}}{\mathrm{3}}\:{and} \\ $$$${f}\left({x}\right)=\left(\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\right)\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)={B}\left({t}\right)\:\Rightarrow \\ $$$${B}\left({t}\right)=\frac{\left(\mathrm{3}{t}−\mathrm{1}\right)\left(\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{1}−\mathrm{3}{t}\right)\left(\mathrm{1}+\mathrm{3}{t}\right)}\:\:\Rightarrow{B}\left({t}\right)=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{3}{t}+\mathrm{1}}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\frac{\mathrm{1}}{\mathrm{3}}} \:\:{B}\left({t}\right)\:=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\frac{\mathrm{2}}{\mathrm{9}}+\mathrm{1}}{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{11}}{\mathrm{9}}\right)\:=−\frac{\mathrm{11}}{\mathrm{54}} \\ $$
Commented by abdomathmax last updated on 14/Mar/20
sir john your answer is false...
$${sir}\:{john}\:{your}\:{answer}\:{is}\:{false}… \\ $$
Commented by jagoll last updated on 14/Mar/20
lim_(x→∞) ((x^2 [((x−1)/(9x+2)) −(1/9)])/(x [((√(x−1))/( (√(9x+2)))) +(1/3)] )) = lim_(x→∞) ((x [((9x−9−9x−2)/(9(9x+2)))])/((3(√(x−1))+(√(9x+2)))/(3(√(9x+2))))) = (1/3)× lim_(x→∞) ((−11x)/(9x+2)) × ((√(9x+2))/( (√(9x−9))+(√(9x+2)))) = (1/3)×lim_(x→∞) ((−11x)/( (√(81x^2 −63x−18))+(√(81x^2 +36x+4)))) = (1/3)×((−11)/(18)) = −((11)/(54))
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\left[\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{9}}\right]}{{x}\:\left[\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{\mathrm{9}{x}+\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{3}}\right]\:}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:\left[\frac{\mathrm{9}{x}−\mathrm{9}−\mathrm{9}{x}−\mathrm{2}}{\mathrm{9}\left(\mathrm{9}{x}+\mathrm{2}\right)}\right]}{\frac{\mathrm{3}\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{9}{x}+\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{9}{x}+\mathrm{2}}}}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\mathrm{11}{x}}{\mathrm{9}{x}+\mathrm{2}}\:×\:\frac{\sqrt{\mathrm{9}{x}+\mathrm{2}}}{\:\sqrt{\mathrm{9}{x}−\mathrm{9}}+\sqrt{\mathrm{9}{x}+\mathrm{2}}}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\mathrm{11}{x}}{\:\sqrt{\mathrm{81}{x}^{\mathrm{2}} −\mathrm{63}{x}−\mathrm{18}}+\sqrt{\mathrm{81}{x}^{\mathrm{2}} +\mathrm{36}{x}+\mathrm{4}}}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\frac{−\mathrm{11}}{\mathrm{18}}\:=\:−\frac{\mathrm{11}}{\mathrm{54}} \\ $$$$ \\ $$
Commented by jagoll last updated on 14/Mar/20
small error sir in line 2
$$\mathrm{small}\:\mathrm{error}\:\mathrm{sir}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2} \\ $$
Commented by john santu last updated on 15/Mar/20
you are right
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$

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