My question is about e) $\displaystyle x*ln(x)$. So to find the approximation. It is said:

$\displaystyle x*ln(x) = (1+h)ln(1+h) \approx (1+h)(h-\frac{h^2}{2}) \approx h + \frac{h^2}{2}$

My question is about $\displaystyle (h-\frac{h^2}{2})$. I thought I studied my approximations manual fairly in depth however I don't recall if anything was said about ln or anything about $\displaystyle (h-\frac{h^2}{2})$? I am wondering why ln(1+h) seems to be approximately equal to $\displaystyle (h-\frac{h^2}{2})$ or should I just accept that it is?

Thanks for any responses...