# Math Help - L'Hopital's Rule?

1. ## L'Hopital's Rule?

Can I get an answer using L'Hopital's rule here?

Lim x --->2 $\frac{x^2-4}{ln(x^2-3)}$

I know that they'll $\frac{0}{0}$, but I just wasn't sure. Can I get some help to work through this one?

2. $\lim_{x\to 2 } \frac{x^2-4}{\ln(x^2-3)}$

$\frac{2^2-4}{\ln(2^2-3)}$

$\frac{4-4}{\ln(1)}$

$\frac{0}{0}$

Yes, you can use L'hopital here.

3. ## L'hôpital

You will have no problem using l'hôpital there, the only problem with l'hôpital could be that you get nowhere by using derivatives, but this will not be the case

4. $\frac{2x}{\frac{1}{x^2-3}}$

plugging in 2

$\frac{4}{1/-1}$ so -4?

5. Hello, XIII13Thirteen!

$\lim_{x\to2}\frac{x^2-4}{\ln(x^2-3)}$
Be careful . . .

The derivative of $x^2-4$ is: . $2x$

The derivative of $\ln(x^2-3)$ is: . $\frac{2x}{x^2-3}$

So we have: . $\lim_{x\to2}\left(\frac{2x}{\frac{2x}{x^2-3}}\right) \;=\;\lim_{x\to2}(x^2-3) \;=\;\boxed{1}$

6. Originally Posted by Soroban
Hello, XIII13Thirteen!

Be careful . . .

Thanks, I forgot about the chain rule in the derivative of the denominator