Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ actually exist? I have seen questions of this type being solved as follows :
$\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this expression be $y$. So
$$\sqrt{x+y} = y \implies x+y = y^2 \implies y^2-y-x=0$$
Using the quadratic formula, we obtain the value of $y$ as :
$$\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-x)}}{2(1)} = \dfrac{1\pm\sqrt{1+4x}}{2} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$$
This has been used to solve $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ in my Mathematics textbook.
Now, this method would work perfectly, assuming that a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exists. If a value for this expression does not exist, this would be similar to Numberphile's popular $\displaystyle\sum_{n=1}^\infty n = \dfrac{-1}{12}$ which is doubtlessly wrong.
So, does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exist? Why/why not?
Thanks!

Kitchen's Calculus of One Variable gives a rigorous solution  although is not exactly the same question (Exercise 8, Section 3-3).
To show that $\sqrt{x}$, $\sqrt{x+\sqrt{x}}, \cdots $ converges to $\frac{1+\sqrt{1+4x}}{2}$, let the nth term be $a_n$. Define $h_n=\frac{1+\sqrt{1+4x}}{2}-a_n$. You can prove by induction that
$$0<h_n\le \frac{\frac{1+\sqrt{1+4x}}{2}-\sqrt{x}}{(\frac{1+\sqrt{1+4x}}{2}+\sqrt{x})^{n-1}}.$$
Now it is easy to see that $h_n\rightarrow 0$ by pinching  since the denominator above is greater than 1 when $x>0$ .
