No matter how I look at this limit, it looks like it should evaluate to $0$. Am I missing some rule about limits with differentials?
The limit is from page 19 of the third edition of Viscous Fluid Flow by Frank White.
$$d\alpha = \lim_{dt->0}\left(\arctan\frac{\frac{\partial v}{\partial x}dxdt}{dx+\frac{\partial u}{\partial x}dxdt}\right) = \frac{\partial v}{\partial x}dt$$
My best guess is that the author is assuming $dx >> \frac{\partial u}{\partial x}$ and then using the small-angle approximation, but that is not mentioned in the text.
Any help is appreciated.
There's not so much a typo as there is a seeming abuse of notation. It seems that White is using $$\lim_{dt->0}\left(\arctan\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right) = \frac{\partial v}{\partial x}dt$$ to mean that $$\arctan\left(\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right)\sim \frac{\partial v}{\partial x}dt\quad\text{ as } dt\to 0.$$ This is relatively straightforward to show. First, rewrite \begin{align} \arctan\left(\frac{\frac{\partial v}{\partial x}dx\,dt}{dx+\frac{\partial u}{\partial x}dx\,dt}\right) = \arctan\left(\frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt}\right). \end{align}
Then, using $$\tan(\theta) \sim \theta\quad\text{ as }\theta \to 0 \implies \arctan(\theta) \sim \theta\quad\text{ as }\theta \to 0,$$ as well as
$$\frac{ax}{1+bx}\sim ax\quad\text{ as }x\to 0, $$ we have that
\begin{alignat}{2} \arctan\left(\frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt}\right) &\sim \frac{\frac{\partial v}{\partial x}dt}{1+\frac{\partial u}{\partial x}dt} &&\quad\text{ as }dt\to 0\\ &\sim \frac{\partial v}{\partial x}dt&&\quad\text{ as } dt\to 0. \end{alignat}