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Under what conditions does the following expression hold?

$$\frac{a+x}{(a+2x)^2} \approx \frac{1}{x}, x>0, a>0$$

I was reading a paper, and the paper pointed out that, when $a \ll x$, the above approximation holds. However, wouldn't that give $\frac{1}{4x}$ as the approximation instead?

Full description in the paper (symbols transformed to general variable names):

$$ f(x)=\frac{x}{a+x_0+x}, a,x_0, x>0$$

Now, get the Taylor expansion to the first order at point $x=x_0$ to get the approximation:

$$ f(x) \propto \frac{x-x_0}{x_0} f(x_0)$$ when $x_0 \gg a$.

I managed to get expand the Taylor series of $f(x)$ to the 1st order and simplify it to get the first expression in this post, but $x_0 \gg a$ doesn't seem to be sufficient to make the approximation. This could just be one of a bag of usual approximations that engineers are very familiar with. I would be more than happy if someone could point me to a comprehensive list of those.

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You are correct, if $\frac{a}{x}\to 0$, then $\frac{a+x}{(a+2x)^2}\to\frac{1}{4x}$, not $\frac{1}{x}$.

Maybe the paper was looking just at the asymptotics, i.e. saying that $\frac{a+x}{(a+2x)^2}=\Theta(\frac{1}{x})$ for $a\ll x$?

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    I found that they introduced a arbitrary constant for proportionality. That can, then, take 1/4 into account. No idea why it didn't occur to me yesterday. Thanks! – 2017-02-28