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&lt;&lt;\ell\,,\ then \sqrt{x^2 - \ell^2}\ is undefined.

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

 

[MahaX]

If x&lt;&lt;\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!