Approximating an expression for a potential In a problem which I was doing, I came across an expression for the potential $V$ of a system as follows $$V = k\left(\frac{1}{l - x} + \frac{1}{l + x}\right)\tag{1}\label{1}$$ where $k$ is a constant, $l$ and $x$ are distances and $l \gg x$. Now I went to find an approximate expression $$V = k\left(\frac{(l + x) + (l - x)}{l^2 - x^2}\right)$$ and reasoning that since $l \gg x$, $l^2 - x^2 \approx l^2$ and thus $$V \approx \frac{2k}{l} \tag{2}\label{2}$$ but this turns out to be wrong as the potential is expected to for an harmonic oscillator and thus propotional to $x^2$. The right way to approximate is \begin{align} V & = \frac{k}{l}\left(\frac{1}{1 - x / l} + \frac{1}{1 + x / l}\right) \\ & \approx \frac{k}{l}\left(\left(1 + \frac{x}{l} + \frac{x^2}{l^2}\right) + \left(1 - \frac{x}{l} + \frac{x^2}{l^2}\right)\right) \\ & \approx \frac{k}{l}\left(2 + \frac{2x^2}{l^2}\right) \end{align}
Ignoring the constant $2$ as I'm concerned about the differences in the potential, I get $$ V \approx \frac{2kx^2}{l^3} \label{3}\tag{3}$$ which is correct.
What mistake did I do in my approximation method?

We have
$$
x^2 \ll l^2
\implies \frac{2x^2}{l^2}
         \ll      2
\implies \frac{2x^2}{l^2} + 2
         \approx  2
$$
So,
$$
V
\approx  \frac{k}{l} \bigg(2 + \frac{2x^2}{l^2}\bigg)
\approx 2\frac{k}{l}
$$
