I came across an interesting approximation today in an indian olympiad paper (page 6). It essentially boils down to:
$$V(x)=\frac{k}{\sqrt{x^2+4/3}}+\frac{k}{\sqrt{(x+1)^2+1/3}}+\frac{k}{\sqrt{(x-1)^2+1/3}}$$ $$\approx k\sqrt{\frac 34}(3+\frac {9}{16}x^2)$$
Even before I got down to attempt to solve this step, I realized a problem. expression 1 decreases with $x$, since $x$ is in the denominator of all three terms in it. But expression 2 increases with $x$, since $x$ is in the numerator. But both are supposed to be approximately equal. How do we justify this?
While it is true that for large $x$, the left side tends to $0$ and the right side tends to $\infty$, these are approximations for small $x$. As Daniel Fischer notes, for small $x$, we need to apply a bit more care.
The first three terms in the Taylor Series for $\frac1{\sqrt{1+x}}$ are $$ (1+x)^{-1/2}=1\color{#090}{-\tfrac12x}\color{#C00}{+\tfrac38x^2}\color{#00F}{-\tfrac5{16}x^3}+O\!\left(x^4\right)\tag1 $$ Thus, $$ \begin{align} \left(\tfrac43+x^2\right)^{-1/2} &=\tfrac{\sqrt3}2\left(1+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{-\tfrac38x^2}\color{#C00}{+O\!\left(x^4\right)}\right)\tag2 \end{align} $$ and $$ \begin{align} \left(\tfrac13+(x+1)^2\right)^{-1/2} &=\left(\tfrac43+2x+x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1+\tfrac32x+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{-\tfrac34x-\tfrac38x^2}\color{#C00}{+\tfrac{27}{32}x^2+\tfrac{27}{32}x^3}\color{#00F}{-\tfrac{135}{128}z^3}+O\!\left(x^4\right)\right)\\ &=\tfrac{\sqrt3}2\left(1-\tfrac34x+\tfrac{15}{32}x^2-\tfrac{27}{128}x^3+O\!\left(x^4\right)\right)\tag3 \end{align} $$ and $$ \begin{align} \left(\tfrac13+(x-1)^2\right)^{-1/2} &=\left(\tfrac43-2x+x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1-\tfrac32x+\tfrac34x^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(1\color{#090}{+\tfrac34x-\tfrac38x^2}\color{#C00}{+\tfrac{27}{32}x^2-\tfrac{27}{32}x^3}\color{#00F}{+\tfrac{135}{128}z^3}+O\!\left(x^4\right)\right)\\ &=\tfrac{\sqrt3}2\left(1+\tfrac34x+\tfrac{15}{32}x^2+\tfrac{27}{128}x^3+O\!\left(x^4\right)\right)\tag3 \end{align} $$ Adding, we get $$ \begin{align} &\left(\tfrac43+x^2\right)^{-1/2}+\left(\tfrac13+(x+1)^2\right)^{-1/2}+\left(\tfrac13+(x-1)^2\right)^{-1/2}\\ &=\tfrac{\sqrt3}2\left(3+\tfrac9{16}x^2+O\!\left(x^4\right)\right)\tag4 \end{align} $$ Therefore, we get an approximation accurate to $O\!\left(x^4\right)$, which is why the plots match so closely, as noted by Claude Leibovici.
Because this is an even function, there will only be terms in the expansion with even exponents of $x$. This is why the terms with $x$ and $x^3$ vanish.