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I tried to figure out how the result in the following limit was obtained but I couldn't.
So how was this limit calculated?

$$ \lim \limits_{dt \to 0} \Big(\tan^{-1}{\frac{\frac{\partial v}{\partial x}\,dx\, dt}{dx+\frac{\partial u}{\partial x}dx\,dt}}\Big) = \frac{\partial v}{\partial x}dt$$

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  • $\begingroup$ All those partial and non-partial differential terms in the (apparently) argument of the arctangent look very odd. What is this, from where does this come? $\endgroup$ – DonAntonio Mar 28 '16 at 6:58
  • $\begingroup$ The derivation of angular speed (or vorticity) of an infinitesimal fluid element. Related to this question. $\endgroup$ – Algo Mar 28 '16 at 7:01
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    $\begingroup$ In fact, consider the 2nd order element in the fraction as zero. Then there is a cancellation that gives exactly the RHS, the reason being that in the vicinity of 0, $tan^{-1} (u)\approx u $ $\endgroup$ – Jean Marie Mar 28 '16 at 9:18
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Rewritten in a more "conventional" form, the question is about finding the following limit $$ \lim_{\epsilon\to 0}\frac{1}{\epsilon}\arctan\left[\frac{A \epsilon}{1+B\epsilon}\right]=A\ , $$ where $$ A=\frac{\partial v}{\partial x}\qquad B=\frac{\partial u}{\partial x}\ . $$ Note that 1) writing a 'limit' for $dt\to 0$ is not compatible with the right hand side being still a function of $dt$, and 2) the $dx$ is immaterial, as it gets cancelled between numerator and denominator. The limit above can be easily computed by expanding the $\arctan$ up to leading order in $\epsilon$ as $\arctan(A\epsilon)\sim A\epsilon$ for $\epsilon\to 0$.

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