A) 0
B) 1
C) 2
D) 1/2
Correct Answer: B
Solution :
Let\[y=\frac{x}{x+\frac{\sqrt[3]{x}}{x+\sqrt[3]{x}...\,\,\text{infinity}}}\] \[=\frac{x}{x+\frac{1}{{{x}^{2/3}}}\cdot \frac{x}{x+\sqrt[3]{x}...\infty }}=\frac{x}{x+\frac{y}{{{x}^{2/3}}}}\] \[\Rightarrow \]\[y=\frac{{{x}^{5/3}}}{{{x}^{5/3}}+y}\] \[\Rightarrow \]\[{{y}^{2}}+({{x}^{5/3}})\cdot y-{{x}^{5/3}}=0\] \[\therefore \]\[y=\frac{-{{x}^{5/3}}\pm \sqrt{{{x}^{10/3}}+4{{x}^{5/3}}}}{2}\] \[=\frac{-{{x}^{5/3}}+\sqrt{{{x}^{10/3}}+4{{x}^{5/3}}}}{2}\] \[(\because \,\,y>0)\] \[=\frac{4{{x}^{5/3}}}{2(\sqrt{{{x}^{10/3}}+4{{x}^{5/3}}+{{x}^{5/3}})}}=\frac{2}{\sqrt{\left( 1+\frac{4}{{{x}^{5/3}}} \right)+1}}\] \[\therefore \]\[\underset{x\to \infty }{\mathop{\lim }}\,y=\frac{2}{\sqrt{1+0}+1}\] \[=\frac{2}{2}=1\]
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