Limiting Fractional Infinities...

Let $\underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right) \quad = \quad \frac { x }{ x+\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } .\frac { x }{ x\quad +\quad \sqrt [ 3 ]{ x } .........\infty } }$

$\therefore \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { x }{ x+\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } .(y) }$

$\therefore \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { { x }^{ \frac { 5 }{ 3 } } }{ { x }^{ \frac { 5 }{ 3 } }\quad +\quad y }$

$\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad { y }^{ 2 }\quad +\quad \left( { x }^{ \frac { 5 }{ 3 } } \right) y\quad -\quad { x }^{ \frac { 5 }{ 3 } }\quad =\quad 0$

$\Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { -\left( { x }^{ \frac { 5 }{ 3 } } \right) \quad \pm \quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } }{ 2 }$

$\Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\quad \frac { -{ x }^{ \frac { 5 }{ 3 } }\quad +\quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } }{ 2 } \quad \quad \quad \quad \quad \quad \quad \left( \because \quad y>0 \right)$

$\Rightarrow \quad \quad \quad \quad \quad \quad \quad y\quad =\frac { 4{ .x }^{ \frac { 5 }{ 3 } } }{ 2\left( { x }^{ \frac { 5 }{ 3 } }\quad +\quad \sqrt { { x }^{ \frac { 10 }{ 3 } }\quad +\quad 4.{ x }^{ \frac { 5 }{ 3 } } } \right) } \quad \quad \left( Rationalising\quad by\quad denominator \right)$

$\therefore \quad \quad \quad \quad \quad \quad \quad \quad \quad \underset { x\rightarrow \infty }{ lim } y\quad =\quad \underset { x\rightarrow \infty }{ lim } \frac { 2 }{ \left\{ 1+\sqrt { 1+\frac { 4 }{ { x }^{ \frac { 5 }{ 3 } } } } \right\} }$

$=\quad \frac{2}{1 \quad + \quad 1}$

$\boxed{\underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\sqrt [ 3 ]{ x } ............\infty } } \right) \quad = \quad 1}$

$\underset { x\rightarrow \infty }{ lim } \left( \frac { x }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right)$

$\underset { x\rightarrow \infty }{ lim } \left( \frac { 1 }{ 1+\frac {1}{(x^{1/3} ) x+\frac { \sqrt [ 3 ]{ x } }{ x+\frac { \sqrt [ 3 ]{ x } }{ ..........\infty } } } } \right)$

Thus when x tends to infinity the denominator that is (1 /the cube root equation you can see) will tend to zero

therefore when x tends to infinity the fraction tends to one

Let $y = x + \dfrac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{\cdots}}}$ . Then we have:

$\begin{aligned} y & = x + \frac {\sqrt[3]x}y \\ y^2 - xy - \sqrt[3]x & = 0 \\ \implies y & = \frac {x+\sqrt{x^2+4\sqrt[3]x}}2 \end{aligned}$

Therefore,

$\begin{aligned} L & = \lim_{x \to \infty} \frac x{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{\cdots}}}} \\ & = \lim_{x \to \infty} \frac xy \\ & = \lim_{x \to \infty} \frac {2x}{x+\sqrt{x^2+4\sqrt[3]x}} & \small \color{#3D99F6} \text{blue} \small \color{#3D99F6} \text{Divide up and down by }x \\ & = \lim_{x \to \infty} \frac 2{1+\sqrt{1+4x^{-\frac 53}}} \\ & = \boxed 1 \end{aligned}$