Identify the repeating pattern

The continued fraction is $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\cdots}}}}$

Since the pattern alternates between $3$ and $4$, define $x=3+\frac{1}{y}$ where $y=4+\frac{1}{x}$

Form equations

From $x=3+\frac{1}{y}, \qquad y=4+\frac{1}{x}$ substitute $y$ into the first equation: $x=3+\frac{1}{4+\frac{1}{x}}$

Simplify the expression

$x=3+\frac{1}{\frac{4x+1}{x}}=3+\frac{x}{4x+1}$

Therefore, $x-3=\frac{x}{4x+1}$

Solve the quadratic equation

$\frac{x}{4x+1}=x-3$

Multiply both sides by $(4x+1)$: $x=(x-3)(4x+1)$

Expand: $x=4x^2+x-12x-3$ $x=4x^2-11x-3$

Rearranging, $4x^2-12x-3=0$

Divide by $1$ and solve using quadratic formula: $x=\frac{12\pm\sqrt{144+48}}{8}=\frac{12\pm\sqrt{192}}{8}$ $x=\frac{12\pm 8\sqrt{3}}{8}=\frac{3\pm 2\sqrt{3}}{2}$

So, $x=\frac{3}{2}\pm \sqrt{3}$

Choose the valid value

Since the continued fraction is clearly positive and greater than $3$, we take the positive root: $x=\frac{3}{2}+\sqrt{3}$

Match with the options

$\frac{3}{2}+\sqrt{3}=1.5+\sqrt{3}$

Hence the correct option is A.

Compare with stored answer

Stored correct answer: A

Our derived answer: A

Therefore, they agree.