# How is this limit result proven?

I tried to figure out how the result in the following limit was obtained but I couldn't.
$$\lim \limits_{dt \to 0} \Big(\tan^{-1}{\frac{\frac{\partial v}{\partial x}\,dx\, dt}{dx+\frac{\partial u}{\partial x}dx\,dt}}\Big) = \frac{\partial v}{\partial x}dt$$
• 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, $tan^{-1} (u)\approx u$ – Jean Marie Mar 28 '16 at 9:18
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 $\epsilon\to 0$.