Limit with natural log in the denominator: $\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$ Value of $\displaystyle\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$ 
The answer is given to be $2$. I'd appreciate an explanation.
 A: Since simple substitution of $x:=1$ would yield the indeterminate form $\frac{0}{0}$,
L'Hôpital's rule to the rescue:  
$$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{f'(x)}{g'(x)}$$  
So, take the derivative of the top and the bottom (not the derivative of the top divided by the bottom).  
$$\lim_{x\rightarrow 1}\frac{x^2-1}{\ln x} = \lim_{x\rightarrow 1}\frac{2x}{1/x}=\lim_{x\rightarrow 1}2x^2= 2$$
A: We don't need to rely on L'Hospital's Rule (not that there's anything wrong with using it here).
We only need recall that $\frac{x-1}{x}\le \log x \le x-1$ for $x>0$, with strict inequalities for $x\ne 1$.  
Then, for $x> 1$
$$x+1=\frac{x^2-1}{x-1}<\frac{x^2-1}{\log x}<\frac{x^2-1}{\frac{x-1}{x}}=x(x+1)$$
By the squeeze theorem, we have
$$\lim_{x\to 1^+}\frac{x^2-1}{\log x}=2\tag1$$
An analogous development shows for $0<x<1$ that 
$$x+1=\frac{x^2-1}{x-1}>\frac{x^2-1}{\log x}>\frac{x^2-1}{\frac{x-1}{x}}=x(x+1)$$
whereby the squeeze theorem, we have
$$\lim_{x\to 1^-}\frac{x^2-1}{\log x}=2\tag2$$
Putting together $(1)$ and $(2)$ yields the coveted limit 
$$\lim_{x\to 1}\frac{x^2-1}{\log x}=2$$
as was to be shown!
A: 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
$$\lim _{ x\rightarrow 1 }{ \frac{  (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  } ={ \frac{\lim _{ x\rightarrow 1 }  (x+1)}{\lim _{ x\rightarrow 1 } \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} }  } =\frac{2}{1}=2.$$
A: As the function is of form $\frac{0}{0}$ we can apply LHR
therefore the derivative of function becomes $\frac{2x}{\frac{1}{x}}$
which is equal to 

$2x^2$

now you know $x\to1$
A: Hint : $$\frac{x^2-1}{\log x} = \frac{(x-1)(x+1)}{\log x}= \frac{x+1}{\frac{\log x}{x-1}}$$
And recall this.
A: $$\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 2 }-1 }{ \ln { x }  } =\lim _{ x\rightarrow 1 }{ \frac { x+1 }{ \frac { \ln { x }  }{ x-1 }  }  }  } =\frac { \lim _{ x\rightarrow 1 }{ \left( x+1 \right)  }  }{ \lim _{ x\rightarrow 1 }{ \left( \frac { \ln { x }  }{ x-1 }  \right)  }  } =2$$
