Need help with taylor series. Evaluate the limit
  $$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$


The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.

Now I need to solve the limit by using Taylor Series and this is what i did so far

$$\begin{align*}
f(x) &amp;= 1-x + \ln x = 1 -x + (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4} (x-1)^4 + \ldots \\
g(x) &amp;= 1+\cos πx = 1+\left[ 1+\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6  +\ldots \right] \\
\frac {f(x)}{g(x)} &amp; = \frac{\frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} {2-\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6+\ldots}
\end{align*}$$

I have no idea where to go to solve for $-\frac{1}{π^2}$ now. Please help
Hint. You have, near $x=1$,
$$1-x + \ln x = 1 -x + (x-1) -(x-1)^2/2+ O(x-1)^3$$ 
$$1-x + \ln x = -(x-1)^2/2+ O(x-1)^3$$ 
and
$$1+\cos πx  = 1 -1+\frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$
$$1+\cos πx  = \frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$ thus

$$\frac{1-x + \ln x}{1+ \cos πx} =\frac{-(x-1)^2/2+ O(x-1)^3}{\frac{\pi^2}2 (x-1)^2 + O(x-1)^3}=\frac{-1/2+ O(x-1)}{\pi^2/2  + O(x-1)}=-\frac{1}{\pi^2}+ O(x-1)$$ 
then 


  $$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=-\frac{1}{\pi^2}.$$
You do not need the Tailor expansion but only the first derivative, the rule of L'Hôpital gives us:
$$
\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=\lim\limits_{x \to 1} \frac{\frac{1}{x}-1}{-\pi\,\sin\pi x}=\lim\limits_{x \to 1} \dfrac{\dfrac{1}{x^2}}{-\pi^2\cos\pi x} = -\dfrac{1}{\pi^2}
$$