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lim-x-x-x-x-x-1-




Question Number 213109 by mathlove last updated on 30/Oct/24
lim_(x→∞) ((√(x+(√(x+(√x)))))/( (√(x+1))))=?
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\mathrm{1}}}=? \\ $$
Answered by MrGaster last updated on 30/Oct/24
=((√(x(1+(1/( (√x)))+(1/x))))/( (√(x(1+(1/x))))))  =(((√x)(√(1+(1/( (√x)))+(1/x))))/( (√x)(√(1+(1/x)))))  =((√(1+(1/( (√x)))+(1/x)))/( (√(1+(1/x)))))  lim_(x→∞) ((√(1+(1/( (√x)))+(1/x)))/( (√(1+(1/x)))))  =((√(1+0+0))/( (√(1+0))))  =((√1)/( (√1)))  =1
$$=\frac{\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}\right)}}{\:\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}} \\ $$$$=\frac{\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$=\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$=\frac{\sqrt{\mathrm{1}+\mathrm{0}+\mathrm{0}}}{\:\sqrt{\mathrm{1}+\mathrm{0}}} \\ $$$$=\frac{\sqrt{\mathrm{1}}}{\:\sqrt{\mathrm{1}}} \\ $$$$=\mathrm{1} \\ $$
Answered by efronzo1 last updated on 30/Oct/24
 = (√(lim_(x→∞)  ((x(1+(√((1/x)+(√(1/x^3 ))))))/(x(1+(1/x))))))   = (√( (1/1))) = 1
$$\:=\:\sqrt{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}}+\sqrt{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }}}\right)}{\mathrm{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)}} \\ $$$$\:=\:\sqrt{\:\frac{\mathrm{1}}{\mathrm{1}}}\:=\:\mathrm{1} \\ $$
Answered by Frix last updated on 31/Oct/24
lim_(x→∞)  ((√(x+(√(x+(√x)))))/( (√(x+1)))) =lim_(t→0^+ )  ((√(1+(√(t+(√t^3 )))))/( (√(t+1)))) =1
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\mathrm{1}}}\:=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{t}+\sqrt{{t}^{\mathrm{3}} }}}}{\:\sqrt{{t}+\mathrm{1}}}\:=\mathrm{1} \\ $$

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