Show that $\lim _{x\to \infty }\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right) =1/2$

Question

I don't know how to start. Is it simple algebraic manipulation where,

if, let $a=\sqrt{x+\sqrt{x+\sqrt{x}}} $

and, $b=\sqrt{x}$

the above equation can be manipulated as

$\implies a-b$$.\:\frac{a+b}{a+b}=\frac{a^2-b^2}{a+b}$

giving, $\frac{\sqrt{x+\sqrt{x}}}{\left(\sqrt{x+\sqrt{+x\sqrt{+x}}}+\sqrt{x}\right)\:}$

Now, my mind can't think of any method to solve further.

Answer 1

$\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\dfrac{\sqrt{x}\sqrt{1+\frac{\sqrt{x}}{x}}}{\sqrt{x}\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}=\dfrac{\sqrt{1+\frac{\sqrt{x}}{x}}}{\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}\to\dfrac{1}{2}$

Answer 2

Which going to $$\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{1+\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}}\rightarrow\frac{1}{2}.$$

Answer 3

A change of variable may help in such questions. Let $$ t := \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x}. \tag{1} $$ Make the substitution $\, x = (4n)^{-2}\,$ and simplify to get $$ t = \frac{\sqrt{1+4n\sqrt{1+4n}}-1}{4n}. \tag{2} $$ As $\,x\to\infty\,$ we have $\,n\to 0.\,$ The expansion around $0$ is $$ t = \frac12 + \frac{n}2 - 2n^2 + \frac{11}2n^3 + \cdots. \tag{3} $$ The limit is $\,\frac12.\,$ You can get the same answer by using the simple approximation $\, \sqrt{1+z}\approx 1+\frac{z}2\,$ in equation $(2)$ twice.

Answer 4

$a=\sqrt{x+\sqrt{x+\sqrt{x}}} $.

$a^2 =x+\sqrt{x+\sqrt{x}} $ and $(\sqrt{x}+\frac12)^2 =x+\sqrt{x}+\frac14 $.

$\sqrt{x+\sqrt{x}}^2 =x+\sqrt{x} $ and $(\sqrt{x}+\frac14)^2 =x+\frac12\sqrt{x}+\frac1{16} \lt x+\sqrt{x} $ so $a > \sqrt{x}+\frac12$.

Numerically, it looks like $a < \sqrt{x}+\frac12+\frac1{8\sqrt{x}} $, so lets see if this can be proved.

$(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2 =x + \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12 $, so if $\sqrt{x+\sqrt{x}} \lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12 $ we are done.

But $(\sqrt{x}+\frac12)^2 =x+\sqrt{x}+\frac14 \gt x+\sqrt{x} $ so $\sqrt{x+\sqrt{x}} \lt \sqrt{x}+\frac12 \lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12 =(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2-x $ so $\sqrt{x+\sqrt{x+\sqrt{x}}} \lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}} $.

Therefore $\sqrt{x}+\frac12 \lt \sqrt{x+\sqrt{x+\sqrt{x}}} \lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}} $.

Answer 5

For large values of $x$, the composition of Taylor series of $$y=\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}$$ gives $$y=\frac 12 +\frac {1}{8 x^{1/2}}-\frac {1}{8 x}+\frac {11}{128 x^{3/2}}+O\left(\frac{1}{x^2}\right)$$

Just for the fun, using a pocket calculator for $x=100$, the exact value is $\sqrt{100+\sqrt{110}}-10\approx 0.5113314$ while the above formula gives $\frac{65451}{128000}\approx 0.5113359$.