How can I simplify a nested square roots limit without using l'Hopital's rule?

chemic_23
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Homework Statement


lim2.jpg



Homework Equations





The Attempt at a Solution



I don't have any idea to simplify the function please help me...
 
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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.
 


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


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.
 


Defennder said:
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.

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.
 


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


Timmo said:
l'Hopital gets messy indeed. But, hey, at least it's honest. :-p

So is factoring out the dominant term.
 
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