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Actually, I have the following expression $a1$:

ClearAll["Global`*"]; 
$Assumptions = {c > 0, h > 0};
a1 = (c + Sqrt[c^2 + 4 h^2])/(4 h Sqrt[c^2 + h^2]) // FullSimplify

Both $c$ and $h$ are positive real numbers. However, I want to show by Mathematica that once $c>>h$, i.e., $c$ is much more than $h$, then $a1$ could be simplified in the limit case to the following expression which is easier to handle:

a2 = 1/(2 h) (1 + h^2/(2 c^2)) // FullSimplify

But I cannot incorporate the equivalency $c>>h$ into the command Limit[].

I will be thankful if someone puts some comments to overcome it.

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  • $\begingroup$ Simplify[ Limit[ Simplify[(c + Sqrt[c^2 + 4 h^2])/(4 h Sqrt[c^2 + h^2]) /. c -> h/z], z -> 0], h > 0] $\endgroup$ – Artes May 16 '17 at 10:04
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    $\begingroup$ Or Simplify[#, h > 0] & //@ Limit[(c + Sqrt[c^2 + 4 h^2])/(4 h Sqrt[c^2 + h^2]) /. c -> h/z, z -> 0] $\endgroup$ – Artes May 16 '17 at 10:16
  • $\begingroup$ I think there is a misunderstanding. We should obtain $a2$ by imposing the condition $c>>h$ on $a1$. Is there a way to reach $a2$ from $a1$? $\endgroup$ – Fazlollah May 16 '17 at 10:49
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    $\begingroup$ You can expand in a "small parameter" to desired order: Normal@Series[a1 /. h -> \[Epsilon] c, {\[Epsilon], 0, 2}] /. \[Epsilon] -> h/c // FullSimplify. $\endgroup$ – jkuczm May 16 '17 at 11:09

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