Limit involving square roots

Hi there,

I am trying to see why this function goes to zero at the limit of large n (for constant a):

$\displaystyle \frac{2}{a}(\sqrt{n}\sqrt{n+a}-2n-a)$

With some factoring out, I've arrived at this:

$\displaystyle \frac{2n}{a}(\sqrt{1+\frac{a}{n}}-1-\frac{a}{2n})$

Where the first (root) term in the bracket should go to 1, and the last to 0, thus the bracket should approach zero in the limit. However, for some reason I don't feel quite comfortable with my method and reasoning. Don't I have to worry about the rate the two terms of the product approach their respective limits? Is there a better way to express the function to make this nuance clear?

Thanks,

Chris

Re: Limit involving square roots

Quote:

Originally Posted by

**entropyslave** Hi there,

I am trying to see why this function goes to zero at the limit of large n (for constant a):

$\displaystyle \frac{2}{a}(\sqrt{n}\sqrt{n+a}-2n-a)$

With some factoring out, I've arrived at this:

$\displaystyle \frac{2n}{a}(\sqrt{1+\frac{a}{n}}-1-\frac{a}{2n})$

Where the first (root) term in the bracket should go to 1, and the last to 0, thus the bracket should approach zero in the limit. However, for some reason I don't feel quite comfortable with my method and reasoning. Don't I have to worry about the rate the two terms of the product approach their respective limits? Is there a better way to express the function to make this nuance clear?

You should not feel comfortable! Yes, the bracket approaches 0 but then you are multiplying it by a term that goes to infinity- you have an "indeterminate" of the form $\displaystyle 0(\infty)$.

Instead do what you typically do with problems involving square roots- multiply both numerator and denominator by the "conjugate":

$\displaystyle (\sqrt{n}\sqrt{n+1}- 2n- a)\frac{\sqrt{n}\sqrt{n+1}+ 2n+ a}{\sqrt{n}\sqrt{n+1}+ 2n+ a}= \frac{n(n+1)- (4n^2- 4an+ a^2)}{\sqrt{n}\sqrt{n+1}+ 2n+ a}$

Re: Limit involving square roots

Quote:

Originally Posted by

**HallsofIvy** You should not feel comfortable! Yes, the bracket approaches 0 but then you are multiplying it by a term that goes to infinity- you have an "indeterminate" of the form $\displaystyle 0(\infty)$.

Instead do what you typically do with problems involving square roots- multiply both numerator and denominator by the "conjugate":

$\displaystyle (\sqrt{n}\sqrt{n+1}- 2n- a)\frac{\sqrt{n}\sqrt{n+1}+ 2n+ a}{\sqrt{n}\sqrt{n+1}+ 2n+ a}= \frac{n(n+1)- (4n^2- 4an+ a^2)}{\sqrt{n}\sqrt{n+1}+ 2n+ a}$

I see, thanks for making this clearer to me!

So there was a typo in my original formula - it should of read (with the 2 as part of the first term):

$\displaystyle \frac{1}{a}(2\sqrt{n}\sqrt{n+a}-2n-a)$

So now following your method, I get:

$\displaystyle \frac{1}{a}[2\sqrt{n}\sqrt{n+a}-(2n+a)]\frac{2\sqrt{n}\sqrt{n+a}+(2n+a)}{2\sqrt{n}\sqrt{n +a}+2n+a}=\frac{1}{a}\frac{4n(n+a)-(4n^2+4an+a^2)}{2\sqrt{n}\sqrt{n+a}+(2n+a)}=\frac{-a}{2\sqrt{n}\sqrt{n+a}+2n+a}$

And now this already clearly goes to zero, but I think it is even clearer if I apply a similar factorisation as my original post:

$\displaystyle =\frac{-a}{2n(1+\sqrt{1+\frac{a}{n}}+\frac{a}{2n})}$

Which for large n looks like $\displaystyle -\frac{a}{4n}$

Thanks for your help!