# Sqrt limit question

1. Feb 12, 2009

### 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

??

Last edited: Feb 13, 2009
2. Feb 12, 2009

### Staff: Mentor

Well, 0/0 is not an answer. Put some numbers in and see what you get. That will at least give you an idea of what the limit might be.

3. Feb 13, 2009

### transgalactic

i agree that 0/0 is not an answer
how to solve it?

4. Feb 13, 2009

### HallsofIvy

That looks to me like a candidate for "rationalizing" Write it as
$$\frac{\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)}{1}$$
and multiply both numerator and denominator by
$$\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}\right)$$

You will get
$$\frac{x+ \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}[/itex] Now use the standard "trick" when x is going to infinity: divide both numerator and denominator by the highest power of x, here $\sqrt{x}$, so every x is moved to the denominator: [tex]\frac{\sqrt{1+ \sqrt{1/x}}}{\sqrt{1+ \sqrt{(1/x)+ \sqrt{1/x^2}}}+ 1}$$

As x goes to infinity, each of those fractions goes to 0.