Question

Limit[Sqrt[x + Sqrt[x + Sqrt[x]]] - Sqrt[x], x -> \[Infinity]]，the answer is 1/2 ,but i don't know how to do it in detail

Answer 1

\[\large \lim \limits_{x\to \infty}~ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x}\]

like this ?

Answer 2

yes

Answer 3

try multiplying by a special kind of 1 : \(\large \dfrac{ \sqrt{x+\sqrt{x+\sqrt{x}}} +\sqrt{x}}{ \sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x}}\)

Answer 4

then...?

Answer 5

\[\begin{align}
& \lim \limits_{x\to \infty}~ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} \\ & = \lim \limits_{x\to \infty}~ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} \times \dfrac{ \sqrt{x+\sqrt{x+\sqrt{x}}} +\sqrt{x}}{ \sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x}} \\

& = \lim \limits_{x\to \infty}~ \dfrac{ \left(\sqrt{x+\sqrt{x+\sqrt{x}}}\right)^2 - \left(\sqrt{x}\right)^2}{ \sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x}} \\

& = \lim \limits_{x\to \infty}~ \dfrac{x+\sqrt{x+\sqrt{x}} - x}{ \sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x}} \\

& = \lim \limits_{x\to \infty}~ \dfrac{\sqrt{x+\sqrt{x}} }{ \sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x}}


\end{align}
\]

Answer 6

next divide numerator and denominator by \(\large \sqrt{x}\) so that it becomes clear what terms vanish as x goes to infinity

Answer 7

thank you very much

Answer 8

when u divide \(\sqrt{x}\) top and bottom, you will see that we can treat the order of \(\sqrt{x+ \sqrt{x + \cdots } }\) same as the order of \(\sqrt{x}\) :

\[
\begin{align}


& = \lim \limits_{x\to \infty}~ \dfrac{(\sqrt{x+\sqrt{x}})/\sqrt{x} }{ (\sqrt{x+\sqrt{x+\sqrt{x}}}+ \sqrt{x})/\sqrt{x}} \\

& = \lim \limits_{x\to \infty}~ \dfrac{\sqrt{1 + \mathcal{O(1/x)}} }{ \sqrt{1+\mathcal{O(1/x)}}+ 1} \\

& = \lim \limits_{1/x\to 0 }~ \dfrac{\sqrt{1 + \mathcal{O(1/x)}} }{ \sqrt{1+\mathcal{O(1/x)}}+ 1} \\

& = \dfrac{\sqrt{1 + 0} }{ \sqrt{1+0 }+ 1} \\

& = \dfrac{1 }{ 2} \\

\end{align}
\]