Consider the expression $$V-\left(V+\frac{\partial V}{\partial x} \,dx\right)+p(x,t)\,dx = m \frac{\partial^2 u}{\partial t^2} \, dx$$
Now take the limit as $dx\to 0$ to obtain a PDE.
How can we obtain a PDE by taking $dx\to 0$? I only obtain $0=0$. Would appreciate some explanation.
Well, $V$ cancels $-V$, so you are left with
$$-\frac{\partial V}{\partial x}dx+p(x,t)dx = m\frac{\partial^2 u}{\partial t^2}dx $$
which is an equality of $1$-forms and which leads immediately to
$$-\frac{\partial V}{\partial x} +p(x,t) = m\frac{\partial^2 u}{\partial t^2} .$$