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Consider the function $f(k) = (1 + \sqrt{1 + n k^2})/m$ (note that this is your expression times $k$). We have $$f(0) = \frac{2}{m}$$ $$f''(0) = \frac{n}{m}$$ So the following Mathematica code does the trick: (2 D[f, {k,2}] / f) /. k -> 0 In your case, you would need to define f = (your quantity)*k. Of course, this isn't really a solution by ...