# Expansion of $x \log \left(\frac{l+x}{x} \right)$ about x=0

$$x \log \left(\frac{l+x}{x} \right)=x \log \frac{l}{x} + O(x^2).$$
• Not sure if it helps, but do note that $\frac{l+x}{x} = \frac{l}{x}+\frac{x}{x} = \frac{l}{x} + 1$. But perhaps your Taylor series method already made use of this? Dec 17 '15 at 2:34
$$x \log \left(\frac{l+x}{x} \right)=x \log(l/x) + x \log(1 + x/l) = x \log(l/x) + xO(x) = x \log(l/x) + O(x^2).$$