Is my solution using L'Hospital's rule wrong? 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.
 A: 
I am not sure where else I have made a mistake

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
A: Note that $\ln(x^2-9)=\ln(x-3)+\ln(x+3)$ so the reciprocal of your function is
$$
\frac{\ln(x-3)+\ln(x+3)}{\ln(x-3)}=1+\frac{\ln(x+3)}{\ln(x-3)}
$$
and the limit of the final fraction is $0$, as it is “$3/(-\infty)$”.
Thus both you and your book made some mistake, because the limit is $1$.
I don't know what the book did, but you computed wrong the derivative of the denominator: if $f(x)=\ln(x^2-9)$, then
$$
f'(x)=\frac{2x}{x^2-9}
$$
by the chain rule or by applying $f(x)=\ln(x-3)+\ln(x+3)$ and finding
$$
f'(x)=\frac{1}{x-3}+\frac{1}{x+3}=\dfrac{x+3+x-3}{(x-3)(x+3)}
$$
(but we used here the chain rule again, don't overlook it).
Hence, after applying l'Hôpital, you get
$$
\frac{\dfrac{1}{x-3}}{\dfrac{2x}{x^2-9}}=\frac{1}{x-3}\frac{(x-3)(x+3)}{2x}=\frac{x+3}{2x}
$$
Always do the algebra before going again with l'Hôpital: in this case blind application to the “four story fraction in $\infty/\infty$ form” would lead to nothing.

Just as a curiosity: if you have a limit of the form
$$
\lim_{x\to a}\frac{\ln(f(x))}{\ln(g(x))}
$$
in an indeterminate form, where
$$
\lim_{x\to a}\frac{f'(x)}{g'(x)}=l
$$
exists finite and nonzero and l'Hôpital assumptions apply, then we can apply l'Hôpital to the given limit and we need to compute
$$
\lim_{x\to a}\frac{f'(x)}{g'(x)}\frac{g(x)}{f(x)}=l\frac{1}{l}=1
$$
Therefore, also
$$
\lim_{x\to a}\frac{\ln(f(x))}{\ln(g(x))}=1
$$
A: I would not use L'Hopital's Rule in this case.
I would use the laws of logarithms to rewrite the expression as:
$\frac {\ln(x-3)} {\ln (x-3) + \ln (x+3)} = \frac 1{1 + \frac{\ln(x+3)}{\ln(x-3)} } \to \frac 1{1 + \frac 6{-\infty}} \to \frac 1{1 + 0^-} = 1$
