Evaluating Limit: $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$$

[Maged Saeed]

could you evaluate this limit , show the steps please , :

$$\lim_{x\to\infty} {\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}}$$

 

[MarkFL]

Hello, Maged Saeed! :D

We ask that when people post problems with which they need help, that they show what they have tried, or give their thoughts on how they should begin. This gives our helpers some kind of idea where you are stuck and/or where you may be going astray. Our goal is to actively engage you in the process of solving the problem so that you learn, not simply provide a solution.

So, can you show what you have done so far?

 

[Maged Saeed]

Thanks for your reply Mr MarkFL (:
What I can say is that I have tried to solve it many ways with no success ,,

I have tried to solve using the conjugate way , multiplying root(x+root(x+root(x+root(x)))+root(x))

but I have got no answer ,,

I'm really stuck with it ,,

By the way ,, it is not a homework question!

Thanks again (:

 

[MarkFL]

I think you are on the right track by multiplying by 1 in the form of the conjugate over itself...doing this, I get:

$$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}}$$

Now, what happens when you divide both the numerator and denominator by the numerator?

 

[Maged Saeed]

Actually I would get the following :

$$one \space over \frac{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$$

I don't know what that would lead to ..

(:

 

[MarkFL]

Since $$\frac{1}{\dfrac{b}{a}}=\frac{a}{b}$$, don't we have the same thing?

Try what I suggested above for your next step...

 

[alyafey22]

Hint:

Using the original formula

Take $\sqrt{x}$ as a common factor , then take a proper substitution.

 

[Maged Saeed]

I'm not sure that I understand what you have said Mr MarkFL and Mr Zaid completely , but I think the following answer maybe effective $$\lim_{x \to \infty} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}} $$

Now by squaring the numerator and the denominator I would get the following

$$\lim_{x \to \infty} \frac{x+\sqrt{x+\sqrt{x}}}{2x+2\sqrt{x}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}} $$

Now , dividing by x , for both numerator and denominator then substituting x by infinity , I would get the following

$$\lim_{x \to \infty} \frac{1+0}{2+0} $$

And The limit equal to

$$\frac{1}{2}$$

Am I right ,,
(;

 

[MarkFL]

I was suggesting that you divide every term in the numerator and the denominator by the numerator, to get:

$$\lim_{x\to\infty}\frac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}{x+\sqrt{x+\sqrt{x}}}}+\sqrt{\dfrac{x}{x+\sqrt{x+\sqrt{x}}}}}$$

And you will get the same result you obtained from squaring.

 

[alyafey22]

$${\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}} = \sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}\sqrt{1+\frac{1}{\sqrt{x}}{\sqrt{1+\frac{1}{\sqrt{x}}}}}}-1\right)$$

Use $t = 1/\sqrt{x}$

$$\frac{1}{t}\left(\sqrt{1+t\sqrt{1+t{\sqrt{1+t}}}}-1\right) = \frac{\sqrt{1+t{\sqrt{1+t}}}}{\sqrt{1+t\sqrt{1+t{\sqrt{1+t}}}}+1} = \frac{1}{2} \,\,\, ;t\to 0$$