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?

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- Apr 30th 2008, 09:41 AMXIII13ThirteenL'Hopital's Rule?
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? - Apr 30th 2008, 09:44 AMwingless
$\displaystyle \lim_{x\to 2 } \frac{x^2-4}{\ln(x^2-3)}$

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

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

$\displaystyle \frac{0}{0}$

Yes, you can use L'hopital here. - Apr 30th 2008, 09:51 AMJerryxL'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, 10:07 AMXIII13Thirteen
$\displaystyle \frac{2x}{\frac{1}{x^2-3}}$

plugging in 2

$\displaystyle \frac{4}{1/-1}$ so -4? - Apr 30th 2008, 10:20 AMSoroban
Hello, XIII13Thirteen!

Quote:

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

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

- Apr 30th 2008, 10:24 AMXIII13Thirteen