L'Hopital's Rule?

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April 30th 2008, 09:41 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?
April 30th 2008, 09:44 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.
April 30th 2008, 09:51 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
April 30th 2008, 10:07 AM
XIII13Thirteen

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

plugging in 2

$\frac{4}{1/-1}$ so -4?
April 30th 2008, 10:20 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: . $\lim_{x\to2}\left(\frac{2x}{\frac{2x}{x^2-3}}\right) \;=\;\lim_{x\to2}(x^2-3) \;=\;\boxed{1}$
April 30th 2008, 10:24 AM
XIII13Thirteen

Quote:

Originally Posted by Soroban

Hello, XIII13Thirteen!

Be careful . . .

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

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