limit, challenge problem

[lookagain]

I read this problem in a book or a magazine. It consists of infinitely nested square roots.


The following limit does exist. $\displaystyle \ \$ Find the limit.



$\displaystyle \displaystyle\lim_{x\to 0^{+}} \ \sqrt{x \ + \ \sqrt{x \ + \ \sqrt{x \ + \ \sqrt{x \ + \ ... \ } \ } \ } \ }$

 

[daon2]

In order to treat a limiting value as a number, we must show convergence. I claim the assignment $\displaystyle f(x)=\sqrt{x+\sqrt{x+\cdots}}$ converges for all $\displaystyle x>0$, but I will show for $\displaystyle x<1$, using the monotone convergence theorem.

To have this we need to show that the sequence defined by $\displaystyle a_0(x)=\sqrt{x}, a_n(x) = \sqrt{x+a_{n-1}}$ converges. For short, call them $\displaystyle a_0, a_n$. Well, first $\displaystyle a_n$ is non-decreasing. Certainly $\displaystyle a_0=\sqrt{x} < \sqrt{x+\sqrt{x}} = a_1$. Now, $\displaystyle a_n^2 = x+a_{n-1} \le x+a_{n} = a_{n+1}^2$ and hence $\displaystyle a_n < a_{n+1}$.

Finally, we need to show $\displaystyle a_n$ is bounded. I claim $\displaystyle a_n\le 3$, which is clearly true for $\displaystyle a_0$. Now, suppose $\displaystyle a_{n-1}\le 3$. $\displaystyle a_n = \sqrt{x+a_{n-1}} \le \sqrt{1+3} = 2 < 3$. Therefore $\displaystyle a_n$ converges, and by construction has limit equal to $\displaystyle f(x)$, i.e., $\displaystyle f(x) = \lim_{n\to \infty} a_n$.

The question now comes down to the answer of $\displaystyle \lim_{x\to{0^+}} f(x)$. Using the relation $\displaystyle f(x) = \sqrt{x+f(x)}$, let's examine the equation $\displaystyle f(x)^2 - f(x) - x=0$. As a "quadratic" we have the roots must satisfy:

$\displaystyle f(x) = \dfrac{1}{2}\left(1\pm \sqrt{4x+1}\right)$.

Let us assume first that the "plus" solution is the correct one. Taking the limit as $\displaystyle x\to 0^{+}$, we see that

$\displaystyle \lim_{x\to 0^{+}}f(x) = \lim_{x\to 0^+} \dfrac{1}{2}\left(1+\sqrt{4x+1}\right) = \dfrac{1}{2}(1+1)=1$.

Now we shall see that the limit cannot be zero, by showing the "minus" solution is incorrect, so we pretend it is correct. Assume that $\displaystyle c$ is a number verrrrrrrrry close to, but bigger than, zero. We have that $\displaystyle 4c+1 > 1$, and hence $\displaystyle f(c) = \dfrac{1-\sqrt{4c+1}}{2} < 0$. But that's impossible, since $\displaystyle f(c) \ge a_n(c) \ge \sqrt{c} > 0$.

Therefore

$\displaystyle \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}\sqrt{x+\sqrt{x+\cdots}} = 1$.