The value of $\underset{x\to \infty }{\lim }\left\{\frac{{\displaystyle x}}{{\displaystyle x+\frac{{\displaystyle \sqrt[3]{x}}}{{\displaystyle x+\frac{{\displaystyle \sqrt[3]{x}}}{{\displaystyle x+\sqrt[3]{x}}}\mathrm{....}\infty }}}}\right\}is$

Let  Y = $\frac{x}{x+\frac{\sqrt[3]{x}}{x+\sqrt[3]{x}}\mathrm{....}\infty }$$=\frac{x}{x+\frac{1}{x^{2/3}}\times \frac{x}{x+\sqrt[3]{x\mathrm{...}\infty }}}=\frac{x}{x+\frac{y}{x^{2/3}}}$ $\Rightarrow y=\frac{x^{5/3}}{x^{5/3}+y}\Rightarrow y^2+\left(x^{5/3}\right)y-x^{5/3}=0$ $\therefore \Rightarrow y=\frac{-x^{5/3}\pm \sqrt{x^{10/3}+4x^{5/3}}}{2}$ $=\frac{-x^{5/3}\pm \sqrt{x^{10/3}+4x^{5/3}}}{2}(\because y>0)$ $=\frac{4x^{5/3}}{2\left(\sqrt{(x^{10/3}+4x^{5/3})}+x^{5/3}\right)}=\frac{2}{\sqrt{1+\left(\frac{4}{x^{5/3}}\right)+1}}$$Hence,\underset{x\to \infty }{\lim }=\frac{2}{\sqrt{1+0}+1}=\frac{2}{2}=1$