# Finding expression for ε

1. Oct 27, 2012

### peripatein

Hello,

Is there a better way to find n such that for every n>n0 |sqrt(n^2+3)-sqrt(n^2-1)|<ε?
I have tried reformulating the expression on the left side so that only one of the square roots remains, to not much avail. The only way I could solve it was by squaring both sides. I did get the correct expression for ε for every ε<1, but it does not work for larger ε's, hence, alas, it is wrong.

2. Oct 27, 2012

### Staff: Mentor

I don't see why you would need to worry about $\epsilon$ ≥ 1. In fact, you should probably be able to assume that $\epsilon$ < 1 - the whole goal of this exercise is to show that you can choose n large enough so that the difference of the two square roots is arbitrarily small.

3. Oct 27, 2012

### peripatein

Thank you very much for replying.
n0 >= sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon)
Which not only seems very messy but also yields wrong results for epsilon greater than 1. The question states that epsilon is positive, so I presume my expression must hold for any epsilon, even if greater than 1.
I know that the right answer should probably be 1/(2epsilon), but how may I derive it?

4. Oct 27, 2012

### peripatein

My bad, I meant that the right answer should probably be 2/epsilon!

5. Oct 27, 2012

### Staff: Mentor

If $\epsilon$ is suitably small, then $\epsilon^4$ and $\epsilon^2$ can be neglected. This makes the expression on the right side approximately equal to √16/$2\epsilon$, which gives you the (corrected) result below.

6. Oct 27, 2012

### peripatein

But what if epsilon were equal to 40? In that case epsilon^4 and epsilon^2 may not be neglected and 2/epsilon still yields the right value for n0, whereas sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon) doesn't!

7. Oct 27, 2012

### Staff: Mentor

Why would anyone care? $\epsilon$ is almost always used to represent small (close to 0) numbers.

8. Oct 27, 2012

### HallsofIvy

Staff Emeritus
Since that is a decreasing function of n, and is 2 when n= 1, as soon as $\epsilon$ is larger than 2, you can take $n_0= 1$.