Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\frac{\ln(u)}{u-1}$, then I infer that $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, then I have:
$$\sqrt{3-2} \rightarrow 1
\\ 3^2-9 \rightarrow 0
$$
and so 
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = 1$$
But this is wrong. So I tried another method:
2: 
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = \frac{\ln(\sqrt{x-2})}{(x-3)(x+3)} = \lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{(x-3)} \cdot \lim_{x \rightarrow 3} \frac{1}{x+3} = \frac{1}{6}$$
Which is also wrong.
My questions are: What did I do wrong in each method and how do I solve this?
EDIT:  If $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, and $\ln(x) \rightarrow 0^+$, does this mean that $\frac{0^+}{0^+} \rightarrow 1$?

Just in case you want to see another approach.
Consider $$A=\frac{\ln(\sqrt{x-2})}{x^2-9}=\frac 12\frac{\ln({x-2})}{x^2-9}$$ Now, as Simply Beautiful Art answered, let $x=u+3$ which makes $$A=\frac{\log (1+u)}{2 u (u+6)}$$ Now, use Taylor series around $u=0$ $$\log(1+u)=u-\frac{u^2}{2}+O\left(u^3\right)$$ which makes $$A=\frac{u-\frac{u^2}{2}+O\left(u^3\right)}{2 u (u+6)}=\frac{1-\frac{u}{2}+O\left(u^2\right)}{2 (u+6)}$$ Now, long division $$A=\frac{1}{12}-\frac{u}{18}+O\left(u^2\right)$$ which shows the limit and how it is approached.
