# L'Hopital's Rule?

• Apr 30th 2008, 10:41 AM
XIII13Thirteen
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?
• Apr 30th 2008, 10:44 AM
wingless
$\lim_{x\to 2 } \frac{x^2-4}{\ln(x^2-3)}$

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

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

$\frac{0}{0}$

Yes, you can use L'hopital here.
• Apr 30th 2008, 10:51 AM
Jerryx
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
• Apr 30th 2008, 11:07 AM
XIII13Thirteen
$\frac{2x}{\frac{1}{x^2-3}}$

plugging in 2

$\frac{4}{1/-1}$ so -4?
• Apr 30th 2008, 11:20 AM
Soroban
Hello, XIII13Thirteen!

Quote:

$\lim_{x\to2}\frac{x^2-4}{\ln(x^2-3)}$
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: . $\lim_{x\to2}\left(\frac{2x}{\frac{2x}{x^2-3}}\right) \;=\;\lim_{x\to2}(x^2-3) \;=\;\boxed{1}$

• Apr 30th 2008, 11:24 AM
XIII13Thirteen
Quote:

Originally Posted by Soroban
Hello, XIII13Thirteen!

Be careful . . .

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