Show that $\lim _{x\to \infty }\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right) =1/2$ 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$$.\:\frac{a+b}{a+b}=\frac{a^2-b^2}{a+b}$
giving, $\frac{\sqrt{x+\sqrt{x}}}{\left(\sqrt{x+\sqrt{+x\sqrt{+x}}}+\sqrt{x}\right)\:}$
Now, my mind can't think of any method to solve further.
 A: $\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\dfrac{\sqrt{x}\sqrt{1+\frac{\sqrt{x}}{x}}}{\sqrt{x}\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}=\dfrac{\sqrt{1+\frac{\sqrt{x}}{x}}}{\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}\to\dfrac{1}{2}$
A: Which going to $$\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{1+\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}}\rightarrow\frac{1}{2}.$$
A: A change of variable may help in such questions. Let
$$ 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} $$ The limit is $\,\frac12.\,$
You can get the same answer by using the simple approximation $\, \sqrt{1+z}\approx 1+\frac{z}2\,$
in equation $(2)$ twice.
A: $a=\sqrt{x+\sqrt{x+\sqrt{x}}}
$.
$a^2
=x+\sqrt{x+\sqrt{x}}
$
and
$(\sqrt{x}+\frac12)^2
=x+\sqrt{x}+\frac14
$.
$\sqrt{x+\sqrt{x}}^2
=x+\sqrt{x}
$
and
$(\sqrt{x}+\frac14)^2
=x+\frac12\sqrt{x}+\frac1{16}
\lt x+\sqrt{x}
$
so
$a > \sqrt{x}+\frac12$.
Numerically,
it looks like
$a < \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$,
so lets see
if this can be proved.
$(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2
=x + \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
$,
so if
$\sqrt{x+\sqrt{x}}
\lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
$
we are done.
But
$(\sqrt{x}+\frac12)^2
=x+\sqrt{x}+\frac14
\gt x+\sqrt{x}
$
so
$\sqrt{x+\sqrt{x}}
\lt \sqrt{x}+\frac12
\lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
=(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2-x
$
so
$\sqrt{x+\sqrt{x+\sqrt{x}}}
\lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$.
Therefore
$\sqrt{x}+\frac12
\lt \sqrt{x+\sqrt{x+\sqrt{x}}}
\lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$.
A: For large values of $x$, the composition of Taylor series of $$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 $\frac{65451}{128000}\approx 0.5113359$.
