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Question Number 115122 by bobhans last updated on 23/Sep/20

lim_(x→∞) (√((x^2 +2x)(x^2 +1))) −(√((x^2 +2x)(x^2 +4))) ?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:−\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:? \\ $$

Commented by malwan last updated on 23/Sep/20

can we solve it with lhopital??

$${can}\:{we}\:{solve}\:{it}\:{with}\:{lhopital}?? \\ $$

Answered by john santu last updated on 23/Sep/20

lim_(x→∞) (√(x^2 +2x)) ((√(x^2 +1))−(√(x^2 +4)))= lim_(x→∞) (√(x^2 +2x)) (((−3)/( (√(x^2 +1))+(√(x^2 +4))))) = −3 ×lim_(x→∞) ((x (√(1+(2/x))))/(x ((√(1+(1/x)))+(√(1+(4/x)))))) = −(3/2).

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}\:\left(\frac{−\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\right)\:= \\ $$$$−\mathrm{3}\:×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:\sqrt{\mathrm{1}+\frac{\mathrm{2}}{{x}}}}{{x}\:\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}}}\right)}\:= \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}. \\ $$

Answered by bemath last updated on 23/Sep/20

Answered by Dwaipayan Shikari last updated on 23/Sep/20

lim_(x→∞) x^2 (√((1+(2/x))(1+(1/x^2 )))) −x^2 (√((1+(2/x))(1+(4/x^2 )))) lim_(x→∞) x^2 ((1+(1/x))(1+(1/(2x^2 )))−(1+(1/x))(1+(2/x^2 ))) lim_(x→∞) x^2 ((1+(1/x))((1/(2x^2 ))−(2/x^2 ))) lim_(x→∞) x^2 ((1/(2x^2 ))+(1/(2x^3 ))−(2/x^2 )−(2/x^3 ))=((1/2)−2)=−(3/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}\:−\mathrm{x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}x}^{\mathrm{2}} \left(\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}x}^{\mathrm{2}} \left(\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\left(\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{3}} }−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by Bird last updated on 24/Sep/20

let f(x) =(√((x^2 +2x)(x^2 +1)))−(√((x^2 +2x)(x^2 +4))) f(x)=x^2 (√((1+(2/x)))).(√(1+(1/x^2 )))−x^2 (√(1+(2/x))).(√(1+(4/x^2 ))) f(x)∼x^2 (1+(1/x))(1+(1/(2x^2 ))) −x^2 (1+(1/x))(1+(2/x^2 ))(x→∞) ⇒ f(x)∼(1+(1/x)){x^2 +(1/2)−x^2 −2} f(x)∼(1+(1/x))(−(3/2)) ⇒ lim_(x→∞) f(x) =−(3/2)

$${let}\:{f}\left({x}\right)\:=\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}.\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{2}}{{x}}}.\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$${f}\left({x}\right)\sim{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right) \\ $$$$−{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)\left({x}\rightarrow\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left\{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}−{x}^{\mathrm{2}} −\mathrm{2}\right\} \\ $$$${f}\left({x}\right)\sim\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\infty} {f}\left({x}\right)\:=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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