calculus - Limit with natural log in the denominator: $lim_{xto1}{frac{x^2 - 1}{ln x}}$

Value of $\displaystyle\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$

The answer is given to be $2$. I'd appreciate an explanation.

Answer

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).

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