The limit I have is the following:
$\displaystyle \lim_{x\to3^{+}} \frac{\ln(x-3)}{\ln(x^{2} – 9)}. $
I applied L'Hospital's rule which gave me:
$\displaystyle \lim_{x\to3^{+}} \frac{\frac{1}{x-3}}{\frac{1}{x^{2} – 9}} = \lim_{x\to3^{+}} \frac{1}{x-3} \cdot \frac{(x-3)(x+3)}{1} = \lim_{x\to3^{+}}\frac{(x-3)(x+3)}{(x-3)} = \lim_{x\to3^{+}}x + 3.$
Since it has a limiting value of 3 the answer should be $6$. However when I checked my solution it was wrong, the answer provided was $-\infty$. I have checked on wolfram alpha and my simplification is right but I am not sure where else I have made a mistake. Any help would be greatly appreciated.
Best Answer
observe that the derivative of $\log(x^2-9)$ is wrong, being $\frac{f'}{f}$
$$\frac{d}{dx}\log(x^2-9)=\frac{2x}{x^2-9}$$
EDIT: Applying de l'Hôpital ONCE (and simplifying the resulting expression) you get 1 as result of your limit