Limit with natural log in the denominator: $\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$

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).

$$\lim_{x\rightarrow 1}\frac{x^2-1}{\ln x} = \lim_{x\rightarrow 1}\frac{2x}{1/x}=\lim_{x\rightarrow 1}2x^2= 2$$

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!

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 $$\lim _{ x\rightarrow 1 }{ \frac{ (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } } ={ \frac{\lim _{ x\rightarrow 1 } (x+1)}{\lim _{ x\rightarrow 1 } \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } } =\frac{2}{1}=2.$$