Suppose I have an $R^2 \rightarrow R$ function
$f(x,y) = \frac{1}{ax + b} \qquad a>0,\ x > 0$
and need to find the area under the path
$\mathbf{r}(t) = \begin{pmatrix} x_0 + t(x_f - x_0) \\ y_0 + t(y_f - y_0)\end{pmatrix}$
along the surface spanned by $f$. To solve this I found the formula for the line integral
$\int_a^b f(\mathbf{r})\ ds = \int_a^b f(\mathbf{r}(t))|\mathbf{r}'(t)|\ dt$
which, applied to this problem, would result in
$\int_0^1 \frac{|\mathbf{r}'(t)|}{a(x_0 + t(x_f -x_0))}\ dt = \frac{|\mathbf{r}'(t)|}{a(x_f-x_0)} \ln \frac{ax_f+b}{ax_0+b}$
and $|\mathbf{r}'(t)| = \sqrt{(x_f-x_0)^2 + (y_f-y_0)^2}$
This can obviously not be correct - if integrated over a path where $x_0 = x_f$, we would divide by zero, even though the area under the curve should be finite if $y_f-y_0$ is finite. Does anyone see the mistake in this reasoning?
EDIT: I just found the problem - for $x_0 = x_f$, $a(x_f-x_0) \rightarrow 0$ and $\ln \frac{ax_f+b}{ax_0+b} \rightarrow 0$, therefore we have $\frac{0}{0}$ and have to apply L'Hôpital's rule:
$\lim_{x_0 \rightarrow x_f} \frac{\ln \frac{ax_f+b}{ax_0+b }}{a(x_f-x_0)} = \frac{1}{ax + b}$
where $x = x_0 = x_f$.