$x$ln$x$ quadratic approximation at $x=1$

Question

$x$ln$x$ quadratic approximation at $x=1$

2 ways :

1) quadratic approx whole thing : $(x-1)^2 + \dfrac {(x-1)^2} 2$

This isn't same as
2) split $x$ and take quad approx of ln$x$ at $x=1$

$x((x-1) - \dfrac 1 2 (x-1)^2) = x(2x - \dfrac 3 2)$

Why are they not same?

Answer 1

Change variable: if $x=1+h$, then \begin{align*}x\ln x&=(1+h)\ln(1+h)=(1+h)\Bigl(h-\frac{h^2}2+o(h^2)\Bigr)=h+\frac{h^2}2+o(h^2)\\ &=x-1+\frac{(x-1)^2}2+o\bigl((x-1)^2\bigr)=\frac{x^2-1}2+o\bigl((x-1)^2\bigr). \end{align*}