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=\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}}
$.
