Approximation of $x\log(x/a)$ for $x$ near a I'm trying to see where the approximation 
$$(x-a) + ((x-a)^2)/2a$$ of $x\log(x/a)$ comes from (for x near a).
Might be missing something very trivial but I've already tried the usual expansions unsuccessfully.
Thanks.
 A: We have
$$f(x) = x \log (x/a)$$
and we want to find an approximation of $f$ near $a$. However,
$$f(x) = a\cdot g(x/a)$$
where
$$g(x) = x \log (x).$$
Now, it is enough to find an approximation of $g$ near $a/a = 1$. That is,
$$g(x) = (x-1) + \frac{1}{2}(x-1)^2 + O\big((x-1)^3\big).$$
Now, going back we get
\begin{align}
f(x) &= a \cdot g(x/a) \\
&= a \cdot \left((x/a-1) + \frac{1}{2}(x/a-1)^2 + O\big((x/a-1)^3\big)\right) \\
&= (x-a) + \frac{a^2}{2a}(x/a - 1)^2 + O\left(\frac{a^3}{a^2}(x/a-1)^3\right) \\
&= (x-a) + \frac{1}{2a}(x-a)^2 + O\left(\frac{1}{a^2}(x-a)^3\right)
\end{align}
which is exactly as desired.
I hope this helps $\ddot\smile$
A: Take the Taylor series centered at $a$ for the function
$$f(x) = x \log \left(\frac x a\right) = x \log x - x \log a$$
Since $f(a) = 0$, the constant term is $0$. Next,
$$f'(a) = \log x + x \frac 1 x - \log a \Big|_{x = a} = 1$$
so the linear term is $x - a$. The quadratic term is then derived from
$$f''(a) = \frac 1 x \Big|_{x = a} = \frac 1 a$$
giving
$$\frac{f''(a)}{2!} (x - a)^2 = \frac 1 {2a} (x - a)^2$$
as desired.
