Shouldn't it be infinitely continued?

Since $x=4-\frac{1}{1-\dfrac{1}{4-\dfrac{1}{1-\dfrac{1}{4-\dfrac{1}{1-\dfrac{1}{4-\dfrac{1}{1-\dfrac{1}{x}}}}}}}}$ repeatedly substituting this to the $x$ in the continued fraction leads us to the infinite continued fraction $x=4-\frac{1}{1-\dfrac{1}{\color{#D61F06}{4-\dfrac{1}{1-\dfrac{1}{4-\dfrac{1}{1-\dfrac{1}{4-\dfrac{1}{1-\dfrac{1}{\cdots}}}}}}}}}$ By passing to the limit, we see that this equation is equivalent to $x = 4-\frac{1}{1-\dfrac{1}{\color{#D61F06}{x}}}$ Solving this equation, we get $\begin{aligned} x &= 4-\frac{1}{1-\dfrac{1}{x}} = 4 - \frac{1}{\dfrac{x-1}{x}} = 4 - \frac{x}{x-1} \\ x(x-1) = x^2 - x &= 4(x - 1) - x = 3x - 4 \\ x^2 - 4x + 4 = (x-2)^2 &= 0 \\ x &= \boxed{2} \end{aligned}$

I did it by brute force, starting from the bottom and evaluating the fraction. It took less than a page.

The verification that 2 is the answer can be done the same way, but is far easier.