I was reading about electric field of uniformly charged ring, of radius $R$, on the axis of the ring at the distance $d$ from the center of the ring and I am confused about usage of differentials. It is written that the differential of normal (to the ring) component of electric field is \begin{equation} \label{field} \tag{1} \text{d}E=\frac{1}{4 \pi \epsilon_0}\frac{d}{r^3}\text{d}q\text{,} \end{equation} where $r$ is the distance from ring to the point at which electric field is calculated and $\text{d} q$ is differential of charge i.e. charge on part of the length $\text{d}l$. Since the ring is uniformly charged, $$\text{d} q=\lambda \text{d} l\text{,}$$ where $\lambda$ is linear charge density. So,equation \ref{field} becomes \begin{equation} \tag{2} \label{field2} \text{d}E=\frac{1}{4 \pi \epsilon_0}\frac{d}{r^3}\lambda \text{d}l \end{equation} In equation \ref{field2} the electric field due to part of the length $\text{d}l$ is written as if the whole charge is at the single point and we could use Coulomb's law for point charges.
From the definition of differential of a function $$\Delta E=\text{d}E+o(\text{d} l)\hspace{1cm}(\text{d} l \rightarrow 0)\text{,}$$ where $\Delta E$ is exact contribution of part of length $\text{d} l$. My question is how to prove that $$\Delta E-\frac{1}{4 \pi \epsilon_0}\frac{d}{r^3}\lambda \text{d}l=o(\text{d}l)\text{?}$$
EDIT 1
As @verdelite pointed out, in this example $\Delta E=\text{d}E$. But, what if instead of a ring, there was a rod of length $2 L$? What would be the electric field at point $P$ on the axis which contains the center of the rod and is normal to the rod? Let's say that point P is at distance $d$ from the rod. Now, the differential of charge will, again, be $\text{d} q=\lambda \text{d} l$. What will be linear approximation of $\Delta E$ (I am assuming normal component only)? Will it be $\frac{1}{4\pi\epsilon_0}\frac{\lambda d}{(l^2+d^2)^{\frac{3}{2}}}\text{d}l$ ($l$ goes from $0$ to $L$), and if so, why?
EDIT 2
I think I have a proof for something similar. In reality charges are distributed discretely, so we can model a rod with finite number of charges. Then, in part of the rod with length $\text{d} l$ there are $n$ charged particles with charge $q$. Also, let the vector $\text{d}\vec l$ be a vector which is parallel to the rod and has a length $\text{d}l$. I will call the position of point at which we are calculating the electric field relative to the start of the part of the rod $\vec r$ and position of the start of the part of the rod relative to the positions of charged particles $\vec{h_i}$, where $i$ goes from $1$ to $n$. It is obvious that $\|\vec{h_i}\|\leq\|\text{d}\vec{l}\|$. The electric field is \begin{equation} \tag{3} \label{discrete} \vec{E}=\sum_{i=1}^n \frac{1}{4\pi\epsilon_0}\frac{q}{\|\vec{r}+\vec{h_i}\|^2}(\vec r+\vec{h_i})\text{.} \end{equation} I will call $\vec{E_L}$ the field that would be present if whole charge was at the start of the part of the rod. \begin{equation} \tag{4} \label{linear} \vec{E_L}=\frac{n q}{4 \pi \epsilon_0}\frac{\vec r}{\|r\|^3} \end{equation} If we subtract equation \ref{linear} from equation \ref{discrete}, we get \begin{equation} \tag{5} \label{difference} \vec{E}-\vec{E_L}=\sum_{i=1}^n \frac{q}{4 \pi \epsilon_0}\left(\frac{\vec{r}+\vec{h_i}}{\|\vec{r}+\vec{h_i}\|^3}-\frac{\vec r}{\|\vec r\|^3}\right)\text{.} \end{equation} The norm of difference \ref{difference} is bounded \begin{aligned} \|\vec{E}-\vec{E_L}\|&\leq \left|\frac{q}{4 \pi \epsilon_0}\right|\sum_{i=1}^n \left\|\frac{\vec{r}+\vec{h_i}}{\|\vec{r}+\vec{h_i}\|^3}-\frac{\vec r}{\|\vec r\|^3}\right\|\\&=\left|\frac{q}{4 \pi \epsilon_0}\right|\sum_{i=1}^n\left\|\frac{(\|\vec r\|^3-\|\vec r+\vec{h_i}\|^3)\vec r+\|\vec r\|^3 \vec{h_i}}{\left(\|\vec r\|\|\vec r+\vec{h_i}\|\right)^3}\right\|\\ &\leq \left|\frac{q}{4 \pi \epsilon_0}\right| \sum_{i=1}^n \frac{\|\vec{h_i}\|\left(4+\frac{3}{\|\vec r\|}\|\vec{h_i}\|+\frac{1}{\|\vec r\|^2}\|\vec{h_i}\|^2\right)}{\|\vec r+\vec{h_i}\|^3}\\ &\leq \left|\frac{q n}{4 \pi \epsilon_0}\right| \frac{\|\text{d}\vec{l}\|\left(4+\frac{3}{\|\vec r\|}\|\text{d}\vec{l}\|+\frac{1}{\|\vec r\|^2}\|\text{d}\vec{l}\|^2\right)}{(\|\vec r\|-\|\text{d}\vec l\|)^3}\text{.} \end{aligned} Since the rod is uniformly charged, $n=\lambda \|\text{d}\vec l\|$, where $\lambda$ is a nonnegative constant. Thus, \begin{aligned} \frac{\|\vec{E}-\vec{E_L}\|}{\|\text{d}\vec{l}\|}\leq \left|\frac{q \lambda}{4 \pi \epsilon_0}\right| \frac{\|\text{d}\vec{l}\|^2\left(4+\frac{3}{\|\vec r\|}\|\text{d}\vec{l}\|+\frac{1}{\|\vec r\|^2}\|\text{d}\vec{l}\|^2\right)}{\|\text{d}\vec{l}\|(\|\vec r\|-\|\text{d}\vec l\|)^3}\rightarrow 0\hspace{1cm}(\|\text{d}\vec l\| \rightarrow 0) \end{aligned} So, $\vec{E_L}$ is the linear approximation of $\vec{E}$. Here I have used the following inequality \begin{aligned} |\|\vec r\|^3-\|\vec r+\vec{h_i}\|^3|&=|\|\vec r\|-\|\vec r+\vec{h_i}\|||\|\vec r\|^2+\|\vec r\|\|\vec r+\vec{h_i}\|+\|\vec r+\vec{h_i}\|^2|\\&\leq \|\vec h_i\| |\|\vec r\|^2+\|\vec r\|\|\vec r\|+\|\vec r\|\|\vec{h_i}\|+\|\vec r\|^2+2\|\vec r\|\|\vec{h_i}\|+\|\vec{h_i}\|^2|\\&=\|\vec{h_i}\||3\|\vec r\|^2+3\|\vec r\|\|\vec{h_i}\|+\|\vec{h_i}\|^2|\text{,} \end{aligned} in the first equation I have used $$\|\vec r\|=\|\vec r+\vec{h_i}-\vec{h_i}\|\leq \|\vec r+\vec{h_i}\|+\|\vec{h_i}\| \implies \|\vec r\|-\|\vec r+\vec{h_i}\|\leq \|\vec{h_i}\|\text{,}$$ and $$\|\vec r+\vec{h_i}\|\leq \|\vec r\|+\|\vec {h_i}\| \implies \|\vec r+\vec{h_i}\|-\|\vec r\|\leq\|\vec{h_i}\|\text{,}$$ so, $$(\|\vec r\|-\|\vec r+\vec{h_i}\|\leq \|\vec{h_i}\|)\land(-(\|\vec r\|-\|\vec r+\vec{h_i}\|)\leq \|\vec{h_i}\|)\implies(|\|\vec r\|-\|\vec r+\vec{h_i}\||\leq \|\vec{h_i}\|)\text{.}$$
