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Question Number 114028 by AbhishekBasnet last updated on 16/Sep/20

Commented by bemath last updated on 17/Sep/20

lim_(x→∞) (√(x^2 +x)) −(√x) = ((1−0)/(2.1)) = (1/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +{x}}\:−\sqrt{{x}}\:=\:\frac{\mathrm{1}−\mathrm{0}}{\mathrm{2}.\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Olaf last updated on 16/Sep/20

(√x)((√(x+1))−(√x)) = x((√(1+(1/x)))−1) ∼_∞ x(1+(1/(2x))−1) →_(x→∞) (1/2)

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

Answered by Olaf last updated on 16/Sep/20

(√x)((√(x+1))−(√x)) = (√x)(1/( (√(x+1))+(√x))) ∼_∞ ((√x)/( (√x)+(√x))) = (1/2)

$$\sqrt{{x}}\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}\right)\:=\:\sqrt{{x}}\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}} \\ $$$$\underset{\infty} {\sim}\frac{\sqrt{{x}}}{\:\sqrt{{x}}+\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Dwaipayan Shikari last updated on 16/Sep/20

(√x)(((x+1−x)/( (√(x+1))+(√x))))=((√x)/( (√x)))((1/( (√(1+(1/x)))+(√1))))=(1/2)

$$\sqrt{{x}}\left(\frac{{x}+\mathrm{1}−{x}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}\right)=\frac{\sqrt{{x}}}{\:\sqrt{{x}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 16/Sep/20

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

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

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