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Question-74766

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Question Number 74766 by chess1 last updated on 30/Nov/19
Commented by mathmax by abdo last updated on 30/Nov/19
changement (1/x)=t lead yo lim_(t→+∞) (√(t+(√(t+(√t))))) −(√(t−(√(t+(√t))))) =lim_(t→+∞) g(t) we have (√(t+(√t)))=(√t)×(√(1+(1/( (√t)))))∼(√t)(1+(1/(2(√t)))) =(√t)+(1/2) ⇒(√(t+(√(t+(√t)))))∼(√(t+(√t)+(1/2))) also (√(t−(√(t+(√t)))))∼(√(t−(√t)−(1/2))) and (√(t+(√t)+(1/2)))=(√t)((√(1+(1/( (√t) ))+(1/(2t)))))∼(√t){1+(1/2)((1/( (√t)))+(1/(2t)))} =(√t)+(1/2) +(1/(4(√t))) also (√(t−(√t)−(1/2)))=(√t)((√(1−(1/( (√t)))−(1/(2t))))) ∼(√t){1−(1/2)((1/( (√t)))+(1/(2t)))} =(√t)−(1/2)−(1/(4(√t))) ⇒ f(t) ∼ (√t)+(1/2)+(1/(4(√t)))−(√t)+(1/2) +(1/(4(√t))) =1+(1/(2(√t))) →1 (t→+∞) ⇒ lim_(x→0^+ ) ((√(....))−(√(....)))=1
$${changement}\:\frac{\mathrm{1}}{{x}}={t}\:{lead}\:{yo}\:{lim}_{{t}\rightarrow+\infty} \sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\:−\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}} \\ $$$$={lim}_{{t}\rightarrow+\infty} {g}\left({t}\right)\:\:{we}\:{have}\:\:\sqrt{{t}+\sqrt{{t}}}=\sqrt{{t}}×\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{t}}}}\sim\sqrt{{t}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\right) \\ $$$$=\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}+\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}}\:{also}\:\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}−\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${and}\:\:\sqrt{{t}+\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}}=\sqrt{{t}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{t}}\:}+\frac{\mathrm{1}}{\mathrm{2}{t}}}\right)\sim\sqrt{{t}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{{t}}}+\frac{\mathrm{1}}{\mathrm{2}{t}}\right)\right\} \\ $$$$=\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}\sqrt{{t}}}\:\:{also}\:\sqrt{{t}−\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}}=\sqrt{{t}}\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{t}}}−\frac{\mathrm{1}}{\mathrm{2}{t}}}\right) \\ $$$$\sim\sqrt{{t}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{{t}}}+\frac{\mathrm{1}}{\mathrm{2}{t}}\right)\right\}\:=\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{{t}}}\:\Rightarrow \\ $$$${f}\left({t}\right)\:\sim\:\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{{t}}}−\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}\sqrt{{t}}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\:\rightarrow\mathrm{1}\:\left({t}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\left(\sqrt{….}−\sqrt{….}\right)=\mathrm{1} \\ $$
Commented by chess1 last updated on 30/Nov/19
thanks
$$\mathrm{thanks} \\ $$

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