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Question Number 34921 by rahul 19 last updated on 13/May/18

lim_(x→∞)  { (x/(x+(((x)^(1/3) )/(x+ (((x)^(1/3) )/(x+(((x)^(1/3) )/(......... infinity ))))))))}

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left\{\:\frac{{x}}{{x}+\frac{\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{{x}+\:\frac{\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{{x}+\frac{\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{.........\:{infinity}\:}}}}\right\} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18

let D_r =t=x+(x^(1/3) /t)  t^2 −tx−x^(1/3) =0  t=((x+(√(x^2 −4.1.(−x^(1/3) ))))/2)  =((x+(√(x^2 +4x^(1/3) )))/2)  lim_(x→∞ (/))   ((lim)/(x→∞)) ((x×2)/(x+(√(x^2 +4x^(1/3) ))))  =((lim)/(x→∞ ))((x×2)/(x{1+(√(1+(4/x^(2−(1/3)) )))))  =((lim)/(x→∞ ))(2/({1+(√(1+(4/∞)))))  =1

$${let}\:{D}_{{r}} ={t}={x}+\frac{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{{t}} \\ $$$${t}^{\mathrm{2}} −{tx}−{x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0} \\ $$$${t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\left(−{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}}{\mathrm{2}} \\ $$$$=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\infty\:\frac{}{}} {{l}\mathrm{i}{m}} \\ $$$$\frac{{lim}}{{x}\rightarrow\infty}\:\frac{{x}×\mathrm{2}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }} \\ $$$$=\frac{{lim}}{{x}\rightarrow\infty\:}\frac{{x}×\mathrm{2}}{{x}\left\{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}} }}\right.} \\ $$$$=\frac{{lim}}{{x}\rightarrow\infty\:}\frac{\mathrm{2}}{\left\{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{\infty}}\right.} \\ $$$$=\mathrm{1} \\ $$

Commented by rahul 19 last updated on 13/May/18

Thank you sir.

$${Thank}\:{you}\:{sir}. \\ $$

Commented by NECx last updated on 15/May/18

wow

$${wow} \\ $$

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