# Nested square roots limit

1. Jan 17, 2009

### chemic_23

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 17, 2009

### Dick

Re: limits

It's more awkward to write than hard.
sqrt(x+sqrt(x))=sqrt(x)*sqrt(1+1/sqrt(x)). Now pull a sqrt(x) out of the outer sqrt so you've got sqrt(x+sqrt(x+sqrt(x)))=sqrt(x)*(1+(1/sqrt(x))*sqrt(1+1/sqrt(x))). The denominator is sqrt(x)*sqrt(1+1/x). Now cancel the sqrt(x) on the outside and take the limit. If you can read that I congratulate you. I THINK I got it right.

3. Jan 17, 2009

### chemic_23

Re: limits

The answer is 1... but how did you came up with the equivalent equation for the numerator? :(

4. Jan 17, 2009

### Timmo

5. Jan 17, 2009

### Defennder

Re: limits

It's probably advisable not to use L'hospital because of the nested square roots. Instead, follow what Dick said (I'm hoping I did it the same way he did because I didn't read his post in detail) and start by pulling out all the square roots by making sure that the denominator and numerator share the same square root over the entire expresion.

Then apply that technique inside the nested root. It'll all simplify to something which you can evaluate the limit to.

6. Jan 17, 2009

### Dick

Re: limits

Right. l'Hopital gets messy. But you can write both numerator and denominator as sqrt(x) times something that goes to 1 as x->infinity. Just factor them both as sqrt(x)*something.

7. Jan 17, 2009

### Timmo

Re: limits

l'Hopital gets messy indeed. But, hey, at least it's honest. :tongue2:

8. Jan 17, 2009

### Dick

Re: limits

So is factoring out the dominant term.