Approximating an expression for a potential

Question

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?

Answer 1

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, $$ V \approx \frac{k}{l} \bigg(2 + \frac{2x^2}{l^2}\bigg) \approx 2\frac{k}{l} $$

Answer 2

There is nothing wrong with your first approximation. You got the leading term $2k/l$ correct. You just did not expand out far enough in powers of $x/l$ to see the $2k x^2/l^3$ term. If you were to keep going, you would find a term with $x^4/l^5$ and so on. You would need this if you wanted to study how the oscillation frequency varies with amplitude. How far you need to go depends on what effects you are after.