# Help with an aproximation

1. Nov 5, 2012

### MahaX

I need to show that: $sqrt(x^2 - l^2) - l ≈ {x^2}/{2l}$

2. That should be valid for x << l

3. First I've tried to isolate l in the sqrt, but it got me nowhere. Anyone could show me a simple way to solve this?

2. Nov 5, 2012

### SammyS

Staff Emeritus
If $x<<\ell\,,\$ then $\sqrt{x^2 - \ell^2}\$ is undefined.

Did you mean $\sqrt{\ell^2-x^2}\ ?$

3. Nov 5, 2012

### MahaX

Sorry, I mean $\sqrt{\ell^2+x^2}\$

4. Nov 5, 2012

### Dick

Multiply sqrt(l^2+x^2)-l by (sqrt(l^2+x^2)+l)/(sqrt(l^2+x^2)+l) (which is 1) and reduce the algebra. Think about what that is approximately equal to if x<<l.

5. Nov 6, 2012

Got it.

Thaks!!