<s>[INST] How to find the Taylor series of order $2$ at $x=1$ of the function $f(x)=\frac{x-1}{\ln (x)}.$ How to find the Taylor series of order $2$ at $x=1$ of the function $$f(x)=\frac{x-1}{\ln (x)}.$$
I don't want to use differentiation because I want to be prepared to find the Taylor series for higher orders. 
My Attempt: Write $$f(x)=\frac{1-x}{1-(1+\ln(x))}.$$ Now I consider the auxiliary function $$\frac{1}{1-u(x)}=1+(u(x)-1)+(u(x)-1)^2+o((u(x)-1)^2)$$ and since $1+\ln(x)\to 1$ as $x\to 1$ we have that $$f(x)=(1-x)(1+(x-x^2/2)+(x-x^2/2)^2+o(x^2))$$ $$=1-x^2/2+o(x^2).$$ However this answer is not correct. Where am I going wrong?
 [/INST] $$\begin{align}\ln(1+x)&=x-x^2/2+x^3/3-\cdots\\ \ln(x)&=(x-1)-(x-1)^2/2+(x-1)^3/3\\ \frac1{\ln x}&=\frac{1}{x}\left(\frac1{1-(x-1)/2+(x-1)^2/3-\cdots}\right)\\&=\frac1{x-1}\cdot \left(1+\frac{x-1}2-\frac{(x-1)^2}3+\frac{(x-1)^2}4+O((x-1)^3)\right)\end{align}$$
So $$\frac{x-1}{\ln(x)}=1+\frac{x-1}2-\frac{(x-1)^2}{12}+O((x-1)^3)$$

Here we used the fact that $$\frac1{1-t}=1+t+t^2+\cdots$$
which is valid for $|t|<1$, which is true close to $x=1$ in the example we use ($x-1$ is small in this region).
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