# Graphic proof of the limit = $\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 + …}}}}}}$ [duplicate]

I am trying to prove that: $$\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 + \cdots}}}}}} = 2$$ with the following figure:

Figure

The approach I'm trying to follow is to show that you can evaluate the sum and it will be between, for example, $$2+\sqrt{2}$$ and $$4$$. Also, I see a pattern with the triangles that are formed by the evaluation of the functions and how they decrease in size and are similar.

Any suggestion?

## marked as duplicate by Winther, rtybase, Feng Shao, Xander Henderson, LeucippusOct 10 at 1:10

• Suppose what you have is represented by $x$. Then since it goes on infinitely, we also have $x=x^2-2$. Solve the quadratic equation for $x=2$ (and, well, $x=-1$ but we know that makes no sense). – Andrew Chin Oct 9 at 18:00
• I feel like there should also be a right triangle realization of this, but it's eluding me at the moment. – Cameron Williams Oct 9 at 18:03
• Note that the graphs of $$f_{n+1}(x)=f\circ f_{n}(x)\qquad f_1(x)=\sqrt{2+x}$$ converge to $y=2$ as $n\to \infty$. See here. – clathratus Oct 9 at 18:10
• I don't believe any rigorous proof can be drawn from the drawing alone. However, the drawing can give you a good motivation for a rigorous proof. For example, you can see that the sequence is monotone and bounded, then you can try to prove those facts, use the Monotone Convergence Theorem to conclude that a limit exists, then prove that the limit is what it seems to be. – Stinking Bishop Oct 9 at 18:12

As you follow the legs of the right triangle upwards and to the right, they will eventually approach the intersection point of the two functions $$y = \sqrt{x}$$ and $$y = x-2$$. The limit must be there because the lines are always getting closer to the point at every step, but it is impossible to go beyond that point. To find the intersection point we simply set the functions equal to each other and solve algebraically.
\begin{align} x-2 &= \sqrt{x} \\\\ (x-2)^2 &= x \\\\ x^2 - 4x + 4 &= x \\\\ x^2 - 5x + 4 &= 0 \\\\ (x-4)(x-1) &= 0 \\\\ x =4 \quad &\text{or} \quad x=1 \\\\ \end{align} Eliminating the extraneous solution we have that $$x=4$$, which means our infinite nested radical has a value of $$y = 2$$.