Let's do some calculus! (27)

Let the limit be $L$ and $y = x + \dfrac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x +...}}}$ ; $\implies L = \lim_{x \to \infty} \dfrac xy$ .

Now consider $y$ as follows.

$\begin{aligned} y & = x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x + \frac {\sqrt[3]x}{x +...}}} = x + \frac {\sqrt[3]x}y \end{aligned}$

$\begin{aligned} \implies y^2 -xy - \sqrt[3] x & = 0 & \small \color{#3D99F6}{\text{Solving the quadratic of }y} \end{aligned}$

$\begin{aligned} \implies y & = \frac {x + \sqrt{x^2+4\sqrt[3]x}}2 & \small \color{#3D99F6}{\text{Note that }y > 0} \end{aligned}$

Now, we have:

$\begin{aligned} L & = \lim_{x \to \infty} \frac xy = \lim_{x \to \infty} \frac x{\frac {x + \sqrt{x^2+4\sqrt[3]x}}2} \\ & = \lim_{x \to \infty} \frac {2x}{x + \sqrt{x^2+4\sqrt[3]x}} & \small \color{#3D99F6}{\text{Divide up and down by }x} \\ & = \lim_{x \to \infty} \frac 2{1 + \sqrt{1+\frac 4{x^{5/3}}}} \\ & = \frac 2{1+\sqrt {1+0}} = \boxed{1} \end{aligned}$