Can I get an answer using L'Hopital's rule here?
Lim x --->2 $\displaystyle \frac{x^2-4}{ln(x^2-3)}$
I know that they'll $\displaystyle \frac{0}{0}$, but I just wasn't sure. Can I get some help to work through this one?
Hello, XIII13Thirteen!
Be careful . . .$\displaystyle \lim_{x\to2}\frac{x^2-4}{\ln(x^2-3)}$
The derivative of $\displaystyle x^2-4$ is: .$\displaystyle 2x$
The derivative of $\displaystyle \ln(x^2-3)$ is: .$\displaystyle \frac{2x}{x^2-3}$
So we have: . $\displaystyle \lim_{x\to2}\left(\frac{2x}{\frac{2x}{x^2-3}}\right) \;=\;\lim_{x\to2}(x^2-3) \;=\;\boxed{1}$