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

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  • Thread starter Maged Saeed
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In summary, Mr MarkFL is trying to find the limit for $\sqrt{x}-\sqrt{x}$ but is having difficulty. He suggests trying to find the limit using the conjugate method, then substituting $x$ by infinity. When he does this, he finds that the limit is $1$ and that $t$ approaches zero.
  • #1
Maged Saeed
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could you evaluate this limit , show the steps please , :

$$\lim_{x\to\infty} {\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}}$$
 
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  • #2
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?
 
  • #3
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 (:
 
  • #4
I think you are on the right track by multiplying by 1 in the form of the conjugate over itself...doing this, I get:

\(\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}}\)

Now, what happens when you divide both the numerator and denominator by the numerator?
 
  • #5
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 ..

(:
 
  • #6
Since \(\displaystyle \frac{1}{\dfrac{b}{a}}=\frac{a}{b}\), don't we have the same thing?

Try what I suggested above for your next step...
 
  • #7
Hint:

Using the original formula

Take $\sqrt{x}$ as a common factor , then take a proper substitution.
 
  • #8
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 ,,
(;
 
  • #9
I was suggesting that you divide every term in the numerator and the denominator by the numerator, to get:

\(\displaystyle \lim_{x\to\infty}\frac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}{x+\sqrt{x+\sqrt{x}}}}+\sqrt{\dfrac{x}{x+\sqrt{x+\sqrt{x}}}}}\)

And you will get the same result you obtained from squaring.
 
  • #10
$${\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}} = \sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}\sqrt{1+\frac{1}{\sqrt{x}}{\sqrt{1+\frac{1}{\sqrt{x}}}}}}-1\right)$$

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

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

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

What is the limit of the expression as x approaches infinity?

The limit of the expression as x approaches infinity is equal to 1.

How do you evaluate the limit using algebraic techniques?

To evaluate the limit, first simplify the expression by combining nested radicals. Then, factor out a square root of x from the entire expression. This will leave you with a simpler expression that can be evaluated using algebraic techniques such as factoring or rationalizing the denominator.

Is the limit affected by the value of x?

Yes, the limit is affected by the value of x. As x approaches infinity, the limit approaches 1. However, for smaller values of x, the limit will be slightly less than 1.

Are there any restrictions on the value of x for this limit to exist?

Yes, the value of x must be greater than or equal to 0 for the limit to exist. This is because the expression contains square roots, which are undefined for negative numbers.

How does this limit relate to the concept of nested radicals?

This limit involves nested radicals, which are expressions with multiple levels of square roots. As x approaches infinity, the expression becomes infinitely nested, but it is still possible to evaluate the limit using algebraic techniques. This demonstrates the concept of nested radicals and their relationship to limits.

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