Taylor expansion and Distribution Let $u(x)$ be the step function and $p_u(x)$ be the distribution defined by 
$$\forall \varphi \in D, 
\langle p_u , \varphi  \rangle
= \lim_{\epsilon \to  0} \left (\varphi(0) \ln(\epsilon)  + \int^{+ \infty } _ {\epsilon} \frac{\varphi(x)}{x} dx  \right )$$ 
Using a Taylor expansion of $\varphi(\epsilon)$, Show that  (in the sense of distributions) $$(u(x) ln(x)))'=P_u $$  

From previous post 
part(a) (asked in another post) Show that, In the sense of distributions, we have $\forall \varphi \in D$ 
$$ \langle (u(x) Ln(x))', \varphi \rangle
= \lim_{\epsilon \to 0} \left  (  \varphi(\epsilon)\ln (\epsilon ) + \int^{+ \infty}_{\epsilon} \frac{\varphi(x)}{x} dx  \right )  $$

Attempt
Taylor Expansion  of $\varphi(\epsilon)$
$$\varphi(\epsilon) = \sum^{n}_{k=0} \frac{\varphi^{(n)}(\epsilon)}{n!} (\epsilon- \epsilon_0)^n  $$
Diffirentiating $u(x) ln(x)$ in parts
$$ (u(x) ln(x))'= u'(x)ln(x)+u(x) \frac{1}{x}= \delta(x)ln(x)+ u(x)/x$$
I am guessing that $P_u =\langle u,\varphi  \rangle$
$$\begin{aligned}
<u ,\varphi> = \int^{+\infty}_{0} \varphi(x)dx
             = \int^{\infty}_{0}\sum^{n}_{k=0} \frac{\varphi^{n}(\epsilon)}{n!} (\epsilon- \epsilon_0)^n  d\epsilon
\end{aligned}$$

Kind of lost at this point can't thread the needle appreciate a nudge towards the right direction

Since you already have the "part (a)", this is much simpler than what you are trying to do. In short, we want to "replace" in the limit the $\varphi(\epsilon)$ by $\varphi(0)$. To do this, we only need to prove that $\lim_{\epsilon\to 0}(\varphi(0)\ln(\epsilon) - \varphi(\epsilon)\ln(\epsilon)) = 0$.
In more details, we have $\varphi(\epsilon) = \varphi(0) + \mathcal O(\epsilon)$, so $\varphi(0)\ln(\epsilon) - \varphi(\epsilon)\ln(\epsilon) = \mathcal O(\epsilon)\ln(\epsilon) = o(1)$, which proves the limit claimed just before. Then, according to part (a),
\begin{align*}
\langle(u\ln)',\varphi\rangle
&=\lim_{\epsilon\to 0}\left( \varphi(\epsilon)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x\right)\\
&=\lim_{\epsilon\to 0}\left( \varphi(0)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x + (\varphi(\epsilon)\ln(\epsilon)-\varphi(0)\ln(\epsilon))\right)
\end{align*}
and according to the previous claim, this is equal to
\begin{align*}
\langle(u\ln)',\varphi\rangle
&=\lim_{\epsilon\to 0}\left( \varphi(0)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x\right)+0
\end{align*}
which is by definition $\langle p_u,\varphi\rangle$.
