Following a simplification of expression I am struggling with this expression:

In particular I get stuck with the simplification from the first to the second line. As far as I can see they replace $\text{m$\ell $}=\mu$. Does the new absolute value of $r^2$ represent a new vector, in other words is this being done: $\displaystyle \left| r\right|^2 = \left| r^2-\dfrac{k^2 \ell ^2}{4}\right|$.
 A: It appears to me that they’ve simply observed that
$$\lim_{\ell\to 0}\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|=|\mathbf{r}|^2\;.$$
Thus,
$$\begin{align*}
\lim_{{\ell\to 0}\atop{m\ell=\mu}}-m\left(\frac{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|-\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}\right)&=-\mu\lim_{\ell\to 0}\frac{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|-\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}{\ell\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}\\\\
&=-\mu\lim_{\ell\to 0}\frac1{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}\lim_{\ell\to 0}\frac{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|-\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}{\ell}\\\\
&=-\frac{\mu}{|\mathbf{r}|^2}\lim_{\ell\to 0}\frac{\left|\mathbf{r}+\frac12\ell\mathbf{k}\right|-\left|\mathbf{r}-\frac12\ell\mathbf{k}\right|}{\ell}\;.
\end{align*}$$
