The value of

Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$

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

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

$\Rightarrow(4 x+1)(x-3)=x$

$\Rightarrow 4 x^{2}-12 x+x-3=x$

$\Rightarrow 4 x^{2}-12 x-3=0$

$\mathrm{x}=\frac{12 \pm \sqrt{(12)^{2}+12 \times 4}}{2 \times 4}=\frac{12 \pm \sqrt{12(16)}}{8}$

$=\frac{12 \pm 4 \times 2 \sqrt{3}}{8}=\frac{3 \pm 2 \sqrt{3}}{2}$

$x=\frac{3}{2} \pm \sqrt{3}=1.5 \pm \sqrt{3}$'

But only positive value is accepted

So, $x=1.5+\sqrt{3}$