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$$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$

this look amazingly, square root of zero is $1$.

$$s=\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$$ $$s=\sqrt{n+s}$$

$$s=\frac{ 1\pm \sqrt{1+4n^2}}{2}$$

$$ \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\frac{1\pm \sqrt{1+4n^2}}{2}$$ $$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\lim_{n \to 0}\frac{1\pm \sqrt{1+4n^2}}{2}$$

But there are two answer $0$ and $1$, which is correct?

vito
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    I'm guessing we actually have $\lim_{n\to0^+}$... – abiessu Mar 05 '16 at 15:00
  • There is a difference between $\lim_{x \to 0}$ and actually setting $x=0$. There is a lot of functions which give different values in each case. And I wouldn't recommend using $n$ in this context since $n$ and $m$ usually mean whole numbers – Yuriy S Mar 05 '16 at 15:33

3 Answers3

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As $\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}\ge0$ and $\sqrt{1+4n^2}\ge1$

$\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\dfrac{1+\sqrt{1+4n^2}}2$

lab bhattacharjee
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First, you must prove if the limit exist.

If $s=1$, then $s=\sqrt{n+s}$ means $1=\sqrt{n+1}$

If $s=0$, then $s=\sqrt{n+s}$ means $0=\sqrt{n+0}$

When n=0, both are correct, but the limit is unique, so the limit doesn't exist.

Carlos Bejines
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You have discovered that the limit of a pointwise convergent sequence of functions is not necessarily continuous.

Define, over $[0,\infty)$, the functions $$ f_0(x)=0, \qquad f_{n+1}(x)=\sqrt{x+f_n(x)} $$ So $f_1(x)=\sqrt{x}$, $f_2(x)=\sqrt{x+\sqrt{x}}$, and so on.

Clearly, $f_n(0)=0$, for every $n$, and $f_n(x)$ is continuous. On the other hand, for $x>0$, we have $$ \lim_{n\to\infty}f_n(x)=\frac{1+\sqrt{1+4x}}{2} $$ (see $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$)

So the sequence of functions $f_n(x)$ converges pointwise to the function $$ f(x)=\begin{cases} 0 & \text{if $x=0$}\\[6px] \dfrac{1+\sqrt{1+4x}}{2} & \text{if $x>0$} \end{cases} $$ and this function is not continuous: $$ \lim_{x\to0}\lim_{n\to\infty}f_n(x)= \lim_{x\to0}\frac{1+\sqrt{1+4x}}{2}=1 $$

You can't exchange the limits: $$ \lim_{n\to\infty}\lim_{x\to0}f_n(x)= \lim_{n\to\infty}0=0 $$

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egreg
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