If $$\frac{1}{1 + \frac{1}{1 + \frac{1}{1+ \frac{1}{x}}}} = \frac{5}{8}$$, then what is the value of x?

Let us evaluate the continued fraction from the innermost term and move outward.

Step 1: Start with the innermost denominator.
Let $$A = 1 + \frac{1}{x} = \frac{x+1}{x}$$

Step 2: Move one level out.
Let $$B = 1 + \frac{1}{A} = 1 + \frac{x}{x+1} = \frac{(x+1) + x}{x+1} = \frac{2x+1}{x+1}$$

Step 3: Move to the next level.
Let $$C = 1 + \frac{1}{B} = 1 + \frac{x+1}{2x+1} = \frac{(2x+1) + (x+1)}{2x+1} = \frac{3x+2}{2x+1}$$

Step 4: The given value is the reciprocal of this $$C$$:
$$\frac{1}{C} = \frac{1}{1 + \frac{1}{1 + \frac{1}{1+\frac{1}{x}}}} = \frac{5}{8}$$

So, $$\frac{2x+1}{3x+2} = \frac{5}{8}$$

Step 5: Cross-multiply to solve for $$x$$:
$$8(2x+1) = 5(3x+2)$$
$$16x + 8 = 15x + 10$$
$$16x - 15x = 10 - 8$$
$$x = 2$$

Therefore, the required value of $$x$$ is $$2$$.

Option A which is: 2