How to Approximate sqrt(x^2 - l^2) - l Using x^2/2l for x << l

[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?

 

[SammyS]

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?

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

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

 

[MahaX]

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

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

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

 

[Dick]

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

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.

 

[MahaX]

Got it.

Thaks!