TITLE: Limit with natural log in the denominator: $\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$
QUESTION [6 upvotes]: Value of $\displaystyle\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$ 
The answer is given to be $2$. I'd appreciate an explanation.

REPLY [3 votes]: They told us not to use l'Hospital if not 100% sure.
So for beginners with a little bit of stamina this is a nice method.
You can directly reduce the diverging factor here.
Recall
$$\ln(x) = \ln ((x-1)+1) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^n,$$
therefore
$$\lim _{ x\rightarrow 1 } \frac {  x ^2 -1 }{ \ln ( x)  }=\lim _{ x\rightarrow 1 }{ \frac{  (x+1)(x-1) }{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^n }  } =\lim _{ x\rightarrow 1 }{ \frac{  (x+1)}{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^{n-1} }  } .  $$
By shifting the index we get
$$ \lim _{ x\rightarrow 1 }{ \frac{  (x+1)}{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^{n-1} }  } = \lim _{ x\rightarrow 1 }{ \frac{  (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  }  .$$
Since 
$$ \lim _{ x\rightarrow 1 }{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  = 1,$$
we have
$$\lim _{ x\rightarrow 1 }{ \frac{  (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  } ={ \frac{\lim _{ x\rightarrow 1 }  (x+1)}{\lim _{ x\rightarrow 1 } \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  } =\frac{2}{1}=2.$$