# sqrt limit question..

• February 13th 2009, 03:58 AM
transgalactic
sqrt limit question..
i need to solve this limit
$
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
$

i tried
$
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\\
\lim_{x->\infty}\left (\frac{\frac{1}{\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}} }-\sqrt{x}}{\frac{1}{\sqrt{x}}} \right)
$

but i get 0/0

??
• February 13th 2009, 04:46 AM
Jester
Quote:

Originally Posted by transgalactic
i need to solve this limit
$
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
$

i tried
$
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\\
\lim_{x->\infty}\left (\frac{\frac{1}{\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}} }-\sqrt{x}}{\frac{1}{\sqrt{x}}} \right)
$

but i get 0/0

??

First let $x = \frac{1}{t^2}$, then the limit becomes

$\lim_{t \to 0} \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2} + \frac{1}{t}}} - \frac{1}{t} = \lim_{t \to 0} \frac{\sqrt{1+t\sqrt{t+1}} - 1}{t}$
Then rationalize

$\lim_{t \to 0} \frac{\sqrt{1+t\sqrt{t+1}} - 1}{t} \cdot \frac{\sqrt{1+t\sqrt{t+1}} + 1}{\sqrt{1+t\sqrt{t+1}} + 1}$

After expanding and canceling you get

$\lim_{t \to 0} \frac{\sqrt{t+1}}{\sqrt{1+t\sqrt{t+1}} + 1} = \frac{1}{2}$
• February 13th 2009, 10:43 AM
Soroban
Hello, transgalactic!

We can solve this head-on ... if we have the stamina.

Quote:

$\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$
Multiply top and bottom by the conjugate . . .

$\frac{\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}} {1} \cdot \frac{\sqrt{x+\sqrt{x + \sqrt{x}}} + \sqrt{x}} {\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}$

. . $= \;\frac{(x + \sqrt{x+\sqrt{x}}) - x}{\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}} \;=\;\frac{\sqrt{x+\sqrt{x}}} {\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}}$

Divide top and bottom by $\sqrt{x}$

. . $\frac{ \dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}} {\dfrac{\sqrt{x+\sqrt{x + \sqrt{x}}}}{\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{x}}} \;= \; \frac{\sqrt{\dfrac{x}{x} + \dfrac{\sqrt{x}}{x}}} {\sqrt{\dfrac{x}{x} + \dfrac{\sqrt{x+\sqrt{x}}}{x}} + 1 }$ . $=\;\frac{\sqrt{1 + \dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{\sqrt{x^3}}}} + 1 }$

$\lim_{x\to\infty}\left(\frac{\sqrt{1 + \dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{\sqrt{x^3}}}} + 1 }\right) \;=\;\frac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}} + 1} \;=\;\frac{1}{1+1} \;=\;\boxed{\frac{1}{2}}$

. . . . . . . $\boxed{\begin{array}{c}\text{Kids, don't try this at home.} \\ \text{I'm a trained professional.}\end{array}}$