Consider $\eta(x;\delta) \in C^\infty$ a family of functions parameterized by $\delta$ with domain $x \in [0,1]$ such that,
\begin{align} \lim_{\delta \to 0} \eta(x;\delta) &= 0 \qquad x < R \\\\ \lim_{\delta \to 0} \eta(x; \delta) &= 1 \qquad x > R \end{align}
for some $R \in (0,1)$. Then for any $\epsilon > 0$,
\begin{align} \lim_{\delta \to 0} \int_{R - \epsilon}^{R + \epsilon} \frac{d \eta}{d x} dx = \lim_{\delta \to 0} [\eta]_{R - \epsilon}^{R + \epsilon} = 1 \end{align}
Now consider the differential equation,
\begin{align} \frac{d}{dx} \left( \frac{du}{dx} + \frac{2u}{x} \right) = \frac{d\eta}{dx} \end{align}
with boundary conditions $u(0) = 0$ and $du/dx(1) = 1$. Further $d^2u/dx^2(0) = 0$ for $u$ to be analytic at $x = 0$.
I am interested in understanding the discontinuities induced in $u$ and/or $du/dx$ in the limit $\delta \to 0$.
If I integrate the differential equation across $[R-\epsilon, R+\epsilon]$ under the limit $\delta \to 0$, I get,
\begin{align} \left[ \frac{du}{dx} + \frac{2u}{x} \right]_{R-\epsilon}^{R+\epsilon} = 1 \\\\ \left[ \left[ \frac{du}{dx} \right] \right] (R) + \left[ \left[ \frac{2u}{x} \right] \right] (R) = 1 \end{align}
where I denote by $[[ \cdot ]](R)$ a jump in the corresponding term at $R$.
Now I can solve this equation numerically using $\eta(x;\delta) = 0.5(1 - tanh(\frac{R-x}{\delta}))$. As I take the limit $\delta \to 0$ I observe that only $du/dx$ suffers a discontinuity at $R$. The function $u$ remains continuous.
I am interested in knowing how I can show that for the above differential equation, the term $u$ remains continuous in the limit $\delta \to 0$.