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Question Number 108131 by mathdave last updated on 14/Aug/20

Commented by Her_Majesty last updated on 14/Aug/20

just a try for x→+∞ we have ((x+(√x))/x)=1+(1/( (√x)))→1 ⇒ (√(x+(√(x+(√x)))))∼(√(x+1)) ⇒ the limit=1 or: (√(x+(√x)))∼(1/2)+(√x) when x→+∞ ⇒ (√(x+(√(x+(√x)))))=(√(x+(1/2)+(√x)))∼(√(x+(√x)))∼(1/2)+(√x) and (√(x+1))∼(√x) ⇒ the limit=1

$${just}\:{a}\:{try} \\ $$$${for}\:{x}\rightarrow+\infty\:{we}\:{have}\:\frac{{x}+\sqrt{{x}}}{{x}}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\rightarrow\mathrm{1} \\ $$$$\Rightarrow\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}\sim\sqrt{{x}+\mathrm{1}} \\ $$$$\Rightarrow\:{the}\:{limit}=\mathrm{1} \\ $$$${or}:\:\sqrt{{x}+\sqrt{{x}}}\sim\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}}\:{when}\:{x}\rightarrow+\infty \\ $$$$\Rightarrow\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}=\sqrt{{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}}}\sim\sqrt{{x}+\sqrt{{x}}}\sim\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}} \\ $$$${and}\:\sqrt{{x}+\mathrm{1}}\sim\sqrt{{x}} \\ $$$$\Rightarrow\:{the}\:{limit}=\mathrm{1} \\ $$

Answered by abdomathmax last updated on 14/Aug/20

let f(x)=((√(x+(√(x+(√x)))))/(√(x+1))) ⇒ for x>0 we get f(x) =(((√x).(√(1+(√((x+(√x))/x^2 )))))/((√x)(√(1+(1/x))))) =((√(1+(√((1/x)+(1/(x(√x)))))))/(√(1+(1/x)))) ⇒lim_(x→+∞) f(x) =1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}{\sqrt{\mathrm{x}+\mathrm{1}}}\:\Rightarrow\:\mathrm{for}\:\mathrm{x}>\mathrm{0}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{x}}.\sqrt{\mathrm{1}+\sqrt{\frac{\mathrm{x}+\sqrt{\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} }}}}{\sqrt{\mathrm{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}}}\:=\frac{\sqrt{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}\sqrt{\mathrm{x}}}}}}{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}}} \\ $$$$\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{1} \\ $$$$ \\ $$

Commented by Her_Majesty last updated on 14/Aug/20

great!

$${great}! \\ $$

Commented by abdomathmax last updated on 15/Aug/20

thanks

$$\mathrm{thanks} \\ $$

Answered by bemath last updated on 15/Aug/20

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

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

Answered by 1549442205PVT last updated on 15/Aug/20

Put ((√(x+(√(x+(√x)))))/( (√(x+(√x)))))=y⇒y^2 =((x+(√(x+(√x))))/(x+(√x))) =((1+(√((1/x)+(√(1/x^3 )))))/(1+(√(1/x))))→1when x→∞ ⇒lim_(x→∞) (y^2 )=1⇒lim_(x→∞) y=1 (as y>0)

$$\mathrm{Put}\:\frac{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}{\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}=\mathrm{y}\Rightarrow\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}{\mathrm{x}+\sqrt{\mathrm{x}}} \\ $$$$=\frac{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }}}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}}}}\rightarrow\mathrm{1when}\:\mathrm{x}\rightarrow\infty \\ $$$$\Rightarrow\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)=\mathrm{1}\Rightarrow\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}y}}=\mathrm{1}\:\left(\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{y}}>\mathrm{0}\right) \\ $$

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