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Help with an aproximation

  1. Nov 5, 2012 #1
    I need to show that: [itex] sqrt(x^2 - l^2) - l ≈ {x^2}/{2l} [/itex]

    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. jcsd
  3. Nov 5, 2012 #2


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    If [itex]x<<\ell\,,\ [/itex] then [itex]\sqrt{x^2 - \ell^2}\ [/itex] is undefined.

    Did you mean [itex]\sqrt{\ell^2-x^2}\ ?[/itex]
  4. Nov 5, 2012 #3
    Sorry, I mean [itex]\sqrt{\ell^2+x^2}\ [/itex]
  5. Nov 5, 2012 #4


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    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.
  6. Nov 6, 2012 #5
    Got it.

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